G9 Math Q1-Week 8- Graph Quadratic Function, finding vertex.ppt
1.
SOLVING PROBLEMS INVOLVINGQUADRATIC
INEQUALITIES
A rectangular box is filled with dice. Each die has a
volume of 1cm³ the length of the box is 3cm greater than
its width and its height is 5cm. Suppose the box holds at
most 140 dice, what are the possible dimensions of the
box?
QUADRATIC FUNCTION
y =ax2 + bx + c
Quadratic Term Linear Term Constant Term
What is the linear term of y = 4x2
– 3? 0x
What is the linear term of y = x2
- 5x ? -5x
What is the constant term of y = x2
– 5x? 0
Can the quadratic term be zero? No!
7.
QUADRATIC FUNCTIONS
The graphof a quadratic function is a:
A parabola can open up
or down.
If the parabola opens up,
the lowest point is called
the vertex (minimum).
If the parabola opens down,
the vertex is the highest point
(maximum).
NOTE: if the parabola opens left or right it is not a function!
y
x
Vertex
Vertex
parabola
8.
y = ax2
+bx + c
The parabola will
open down when the
a value is negative.
The parabola will
open up when the a
value is positive.
STANDARD FORM
y
x
The standard form of a quadratic function is:
a > 0
a < 0
9.
y
x
Axis of
Symmetry
AXIS OFSYMMETRY
Parabolas are symmetric.
If we drew a line down
the middle of the
parabola, we could fold
the parabola in half.
We call this line the
Axis of symmetry.
The Axis of symmetry ALWAYS
passes through the vertex.
If we graph one side of
the parabola, we could
REFLECT it over the
Axis of symmetry to
graph the other side.
Find the Axisof symmetry for y = 3x2
– 18x + 7
FINDING THE AXIS OF SYMMETRY
When a quadratic function is in standard form
the equation of the Axis of symmetry
is
y = ax2
+ bx + c,
2
b
a
x
This is best read as …
‘the opposite of b divided by the quantity of 2 times a.’
The Axis of
symmetry is
x =
a = 3 b = -18
12.
FINDING THE VERTEX
TheAxis of symmetry always goes through the
_______. Thus, the Axis of symmetry gives
us the ____________ of the vertex.
STEP 1: Find the Axis of symmetry
Vertex
Find the vertex of y = -2x2
+ 8x - 3
2
b
a
x
a = -2 b = 8
X-coordinate
The x-
coordinate
of the vertex
is 2
13.
FINDING THE VERTEX
STEP1: Find the Axis of symmetry
STEP 2: Substitute the x – value into the original equation
to find the y –coordinate of the vertex.
The
vertex is
(2 , 5)
Find the vertex of y = -2x2
+ 8x - 3
14.
GRAPHING A QUADRATICFUNCTION
There are 3 steps to graphing a parabola in
standard form.
STEP 1: Find the Axis of symmetry using:
STEP 2: Find the vertex
STEP 3: Find two other points and reflect them across
the Axis of symmetry. Then connect the five points
with a smooth curve.
MAKE A TABLE
using x – values close to
the Axis of symmetry.
2
b
a
x
15.
STEP 1: Findthe Axis of
symmetry
( )
4
1
2 2 2
b
x
a
-
= = =
y
x
1
4
2
: 2
x
x
y
Graph
Graphing a Quadratic Function
STEP 2: Find the vertex
Substitute in x = 1 to
find the y – value of the
vertex.
( ) ( )
2
2 1 4 1 1 3
y = - - = -
16.
5
–1
( ) ()
2
2 3 4 3 1 5
y = - - =
STEP 3: Find two other
points and reflect them
across the Axis of
symmetry. Then connect
the five points with a
smooth curve.
y
x
( ) ( )
2
2 2 4 2 1 1
y = - - = -
3
2
y
x
Graphing a Quadratic Function
17.
y
x
Y-intercept of aQuadratic Function
Y-axis
The y-intercept of a
Quadratic function can
Be found when x = 0.
The constant term is always the y- intercept
18.
SOLVING A QUADRATIC
Thenumber of real solutions is at most
two.
No solutions One solution
X = 3
Two solutions
X= -2 or X = 2
The x-intercepts (when y = 0) of a quadratic function
are the solutions to the related quadratic equation.
19.
IDENTIFYING SOLUTIONS
X =0 or X = 2
Find the solutions of 2x - x2
= 0
The solutions of this
quadratic equation can be
found by looking at the
graph of f(x) = 2x – x2
The x-
intercepts(or
Zero’s) of
f(x)= 2x – x2
are the solutions
to 2x - x2
= 0
20.
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Editor's Notes
#6 Ask students “Why is ‘a’ not allowed to be zero? Would the function still be quadratic?
#7 Let students know that in Algebra I we concentrate only on parabolas that are functions; In Algebra II, they will study parabolas that open left or right.
#8 Remind students that if ‘a’ = 0 you would not have a quadratic function.
#11 Discuss with the students that the line of symmetry of a quadratic function (parabola that opens up or down) is always a vertical line, therefore has the equation x =#. Ask “Does this parabola open up or down?
#18 Remind students that x-intercepts are found by setting y = 0 therefore the related equation would be ax2+bx+c=0. Also state that since the highest degree of a quadratic is 2, then there are at most 2 solutions. For the first graph ask “why are there no solutions?”-- there are no solutions because the parabola does not intercept the x-axis. 2nd and 3rd graph ask students to state the solutions. Additional Vocab may be itroduced: The x-intercepts are solutions, zero’s or roots of the equation.
#19 Point out to students that the function can also be written as y = -x2+2x.