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# 2.3 first and second degree functions

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## 2.3 first and second degree functionsPresentation Transcript

• First Degree Functions
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines.
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts,
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept.
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept.
a.2x – 3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept.
(0,-4)
a.2x – 3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept.
(6,0)
(0,-4)
a.2x – 3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept.
(6,0)
(0,-4)
a.2x – 3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
(0,-4)
a.2x – 3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
(0,-4)
a.2x – 3y = 12 b. -3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
a.2x – 3y = 12 b. -3y = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
a.2x – 3y = 12 b. -3y = 12 c. 2x = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
x = 6
a.2x – 3y = 12 b. -3y = 12 c. 2x = 12
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
x = 6
a.2x – 3y = 12 b. -3y = 12 c. 2x = 12
If both x and y are
present, we get a
tilted line.
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
x = 6
a.2x – 3y = 12 b. -3y = 12 c. 2x = 12
If the equation is
y = c, we get a horizontal line.
If both x and y are
present, we get a
tilted line.
• First Degree Functions
The graphs of the equations Ax + By = C are straight
lines. Its easy to graph lines by graphing the x and y
intercepts, i.e. set x = 0 to get y-intercept, and set y = 0 for the x-intercept. If there is only one variable in the equation, we get a vertical or a horizontal line.
(6,0)
y = –4
(0,-4)
x = 6
a.2x – 3y = 12 b. -3y = 12 c. 2x = 12
If the equation is
y = c, we get a horizontal line.
If the equation is
x = c, we get a vertical line.
If both x and y are
present, we get a
tilted line.
• First Degree Functions
Given Ax + By = C with B = 0,
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept,
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12 
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
2
2
3
3
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12
2
2
3
3
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12  y = 0x – 4, so the slope = 0
2
2
3
3
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12  y = 0x – 4, so the slope = 0
c. 2x = 12,
2
2
3
3
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12  y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
2
2
3
3
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12  y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
2
2
3
3
The slope m is also the ratio of the change in the
output vs. the change in the input.
• First Degree Functions
Given Ax + By = C with B = 0, treating x as the input
and y as the output, we may solve for y and put the
equation in a function form: y = f(x) = mx + b,
m is called the slopeand b is the y intercept, and the
form is called the slope-intercept form.
From the examples above,
a. 2x – 3y = 12  y = x – 4, so the slope =
b. -3y = 12  y = 0x – 4, so the slope = 0
c. 2x = 12, the slope is undefined since we can't
solve for y.
2
2
3
3
The slope m is also the ratio of the change in the
output vs. the change in the input. If two points on the
line are given, the slope may be calculated via the
following formula.
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
Δy
m =
Δx
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
Δy
m =
Δx
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
y2 – y1
Δy
m =
=
Δx
x2 – x1
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
y2 – y1
Δy
m =
=
Δx
x2 – x1
rise
=
run
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
m =
=
Δx
x2 – x1
(x1, y1)
rise
=
run
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
Δy=y2–y1=rise
m =
=
Δx
x2 – x1
(x1, y1)
rise
=
run
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
Δy=y2–y1=rise
m =
=
Δx
x2 – x1
(x1, y1)
rise
Δx=x2–x1=run
=
run
(The Greek letter Δ means "the difference".)
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
Δy=y2–y1=rise
m =
=
Δx
x2 – x1
(x1, y1)
rise
Δx=x2–x1=run
=
run
(The Greek letter Δ means "the difference".)
The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1,y1) is a point
on the line, then
y = f(x) = m(x – x1) + y1
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
Δy=y2–y1=rise
m =
=
Δx
x2 – x1
(x1, y1)
rise
Δx=x2–x1=run
=
run
(The Greek letter Δ means "the difference".)
The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1,y1) is a point
on the line, then
y = f(x) = m(x – x1) + y1
First degree functions are also called linear functions
because their graphs are straight lines.
• First Degree Functions
Slope Formula: Let (x1, y1) and (x2, y2) be two points on a line, then the slope
(x2, y2)
y2 – y1
Δy
Δy=y2–y1=rise
m =
=
Δx
x2 – x1
(x1, y1)
rise
Δx=x2–x1=run
=
run
(The Greek letter Δ means "the difference".)
The Point Slope Formula: Let y = f(x) be a first degree equation with slope m, and (x1,y1) is a point
on the line, then
y = f(x) = m(x – x1) + y1
First degree functions are also called linear functions
because their graphs are straight lines.
Linear functions are the most useful important used to approximate other functions.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000,
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points;
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500).
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500). For the linear price function
p(x) = mx+ b, we find the slope
m = (900 – 500)/(0 – 5) = -80.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500). For the linear price function
p(x) = mx+ b, we find the slope
m = (900 – 500)/(0 – 5) = -80.
Hence the price function p(x) = -80(x – 0) + 900,
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500). For the linear price function
p(x) = mx+ b, we find the slope
m = (900 – 500)/(0 – 5) = -80.
Hence the price function p(x) = -80(x – 0) + 900, or
p(x) = -80x + 900.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500). For the linear price function
p(x) = mx+ b, we find the slope
m = (900 – 500)/(0 – 5) = -80.
Hence the price function p(x) = -80(x – 0) + 900, or
p(x) = -80x + 900.
Year 2008 corresponds to x = 8.
• First Degree Functions
Example A: In the year 2000, a cheap laptop cost \$900, in the year 2005, a cheap one cost \$500. Assume the price is linear, find the equation of the price in terms of time. What is the projected price in the year 2008?
Let x = 0 be the year 2000, so the year 2005 is x = 5. Let y = price of a laptop. We have two data points; (0, 900), and (5, 500). For the linear price function
p(x) = mx+ b, we find the slope
m = (900 – 500)/(0 – 5) = -80.
Hence the price function p(x) = -80(x – 0) + 900, or
p(x) = -80x + 900.
Year 2008 corresponds to x = 8.
So p(8) = -80*8+ 900 = \$260 is the projected price.
• First Degree Functions
More Facts on Slopes:
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
2x/3 – 5/2 = y
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
2x/3 – 5/2 = y
Hence the slope of 4x – 2y = 5 is 2/3.
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
2x/3 – 5/2 = y
Hence the slope of 4x – 2y = 5 is 2/3.
Therefore L has slope -3/2.
• First Degree Functions
More Facts on Slopes:
• Parallel lines have the same slope.
• Slopes of perpendicular lines are the negative
reciprocal of each other.
Example B: Find the equation of the line L that
passes through (2, -4) and is perpendicular to
4x – 3y = 5.
Solve for y to find the slope of 4x – 3y = 5
4x – 5 = 3y
2x/3 – 5/2 = y
Hence the slope of 4x – 2y = 5 is 2/3.
Therefore L has slope -3/2. So the equation of L is
y = -3/2(x – 2) + (-4) or y = -3x/2 – 1.
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Below is the graph of the y = f(x) = -x2.
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Below is the graph of the y = f(x) = -x2.
x y
-4
-3
-2
-1
0
1
2
3
4
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Below is the graph of the y = f(x) = -x2.
x y
-4
-3
-2
-1
0 0
1 -1
2 -4
3 -9
4 -16
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Below is the graph of the y = f(x) = -x2.
x y
-4 -16
-3 -9
-2 -4
-1 -1
0 0
1 -1
2 -4
3 -9
4 -16
• Second Degree Functions
Quadratic functions or second degree functions are
functions of the form y = f(x) = ax2 + bx + c, a = 0.
Below is the graph of the y = f(x) = -x2.
x y
-4 -16
-3 -9
-2 -4
-1 -1
0 0
1 -1
2 -4
3 -9
4 -16
• Second Degree Functions
equations are called parabolas.
• Second Degree Functions
equations are called parabolas.
Parabolas describe the paths of thrown objects (or the upside- down paths).
• Second Degree Functions
equations are called parabolas.
Parabolas describe the paths of thrown objects (or the upside- down paths).
• Second Degree Functions
Properties of Parabolas:
• Second Degree Functions
Properties of Parabolas:
I. A parabolas is symmetric to a center line.
• Second Degree Functions
Properties of Parabolas:
I. A parabolas is symmetric to a center line.
II. The highest or lowest point of the parabola sits on the center line.
• Second Degree Functions
Properties of Parabolas:
I. A parabolas is symmetric to a center line.
II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex.
• Second Degree Functions
Properties of Parabolas:
I. A parabolas is symmetric to a center line.
II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex.
III. If the positions of the vertex and another point on the parabola are known, the parabola is determined.
• Second Degree Functions
Properties of Parabolas:
I. A parabolas is symmetric to a center line.
II. The highest or lowest point of the parabola sits on the center line. This point is called the vertex.
III. If the positions of the vertex and another point on the parabola are known, the parabola is determined.
The position of the vertex is given by vertex formula:
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point.
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola.
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
4. Set y=0 to find the x intercept.
-b
2a
• Second Degree Functions
Vertex Formula: The vertex of the
parabola y = ax2 + bx + c is at x =
-b
2a
Moreover, if a > 0, the parabola opens up,
if a < 0 the parabola opens down.
To graph a parabola y = ax2 + bx + c:
Set x = in the equation to find the vertex.
2. Find another point. (Use x=0 to get the y intercept
if its not the vertex.)
3. Locate the reflection across the center line, these
three points form the tip of the parabola. Trace the
parabola.
4. Set y=0 to find the x intercept. If no real solution
exists, there is no x intercept.
-b
2a
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x =
-(-4)
2(1)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2
-(-4)
2(1)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12
-(-4)
2(1)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
(4, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
(4, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
Set y = 0 and get x-int:
x2 – 4x – 12 = 0
(4, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
Set y = 0 and get x-int:
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x = -2, x = 6
(4, -12)
(0, -12)
(2, -16)
• Second Degree Functions
Example C: Graph y = x2 – 4x – 12
Vertex: set x = = 2 then y=22 – 4*2 – 12 = -16,
-(-4)
2(1)
so v=(2, -16).
Another point:
Set x = 0 then y = -12
or (0, -12)
It's reflection across the mid-line is (4, -12)
Set y = 0 and get x-int:
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x = -2, x = 6
(4, -12)
(0, -12)
(2, -16)
Following is an example of maximization via the vertex.
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of times of \$0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x)
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of times of \$0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of times of \$0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
This a 2nd degree equation whose graph is a parabola opens downward.
• Second Degree Functions
Example D: The "Crazy Chicken" can sell 120 whole roasted chicken for \$8 in one day. For every \$0.50 increases in the price, 4 less chicken are sold. Find the price that will maximize the revenue (the sale).
Set x to be the number of \$0.50 increments above the base price of \$8.00.
So the price is (8 + 0.5x) or (8 + x/2).
The number of chicken sold at this price is (120 – 4x).
Hence the revenue, depending on the number of times of \$0.50 price hikes x, is
R(x) = (8 + x/2)(120 – 4x) = 960 + 28x – 2x2.
This a 2nd degree equation whose graph is a parabola opens downward. The vertex is the highest point on the graph where R is the largest.
• Second Degree Functions
R
x
R = 960 + 28x – 2x2.
• Second Degree Functions
The vertex of this parabola is at
x = = 7.
R
-(28)
2(-2)
x
R = 960 + 28x – 2x2.
• Second Degree Functions
The vertex of this parabola is at
x = = 7.
So R(7) = \$1058 gives the maximum revenue.
R
v = (7, 1058)
-(28)
2(-2)
x
R = 960 + 28x – 2x2.
• Second Degree Functions
The vertex of this parabola is at
x = = 7.
So R(7) = \$1058 gives the maximum revenue.
R
v = (7, 1058)
-(28)
2(-2)
More precisely, raise the price x = 7 times to
8 + 7*0.50 = 11.50 per
x
R = 960 + 28x – 2x2.
chicken,
• Second Degree Functions
The vertex of this parabola is at
x = = 7.
So R(7) = \$1058 gives the maximum revenue.
R
v = (7, 1058)
-(28)
2(-2)
More precisely, raise the price x = 7 times to
8 + 7*0.50 = 11.50 per
x
R = 960 + 28x – 2x2.
chicken, and we can sell 120 – 4*7 = 92 per day with the revenue of 11.5 * 92 = 1058.