5 4 the graphs of quadratic equations

Graphs of Quadratic Equations
Frank Ma © 2011

We start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.

Example A. Graph y = –x2

Make a table

Make a table
x
-4
-3
-2
-1
0
1
2
3
4
y

Make a table
x
-4
-3
-2
-1
0
1
2
3
4
y
-16
-9
-4
-1
0
-1
-4
-9
-16

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).

Properties of Parabolas:
Parabolas are symmetric with respect to a center line

The highest or lowest point of the parabola sits on the
center line.

center line. This point is called the vertex.

center line. This point is called the vertex.
-b
Vertex Formula: The vertex of y = ax2 + bx + c is at x = .
2a

One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

Example B. Graph y = x2 – 4x – 12

Vertex: set x = = 2
–(–4)
2(1)

Vertex: set x = = 2
–(–4)
2(1)
Make a table centered at x = 2.
x y
0
1
2
3
4

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
Note the y values are symmetric around the vertex just as the points.

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
(2, -16)

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
(3, -15)
(2, -16)

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
(4, -12)
(3, -15)
(2, -16)

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
(4, -12)
(3, -15)
(1, -15)
(2, -16)

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
–12
–15
–16
–15
–12
(4, -12)
(0, -12)
(3, -15)
(1, -15)
(2, -16)

When graphing parabolas, we must also give the x-intercepts and the y-intercept.

The y-intercept is (0, c) obtained by setting x = 0.

The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c

The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions.

The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex.

The center line is determined by the vertex. Suppose we know another point on the parabola,

The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola.

The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola. There is exactly one parabola that goes through these three points.

(2nd way) To graph a parabola y = ax2 + bx + c.

Set x = in the equation to find the vertex.
-b
2a

2. Find another point, use the y-intercept (0, c) if feasible.
-b
2a

3. Locate its reflection across the center line. These three points form the tip of the parabola.
-b
2a

3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.
-b
2a

4. Set y = 0 and solve to find the x intercept.
-b
2a

-b
2a
Example C. Graph y = –x2 + 2x + 15

-b
2a
The vertex is at x = 1, y = 16

-b
2a
y-intercept is at (0, 15)

-b
2a
Plot its reflection (2, 15)

-b
2a
Draw, set y = 0 to get x-int:
–x2 + 2x + 15 = 0

-b
2a
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0

-b
2a
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5

-b
2a
(1, 16)
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5

-b
2a
(1, 16)
(0, 15)
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5

-b
2a
(1, 16)
(0, 15)
(2, 15)
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5

-b
2a
(1, 16)
(0, 15)
(2, 15)
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5
(-3, 0)
(5, 0)

Finally, we make the observation that given y = ax2 + …,
if a > 0, then the parabola opens upward.

Finally, we make the observation that given y = ax2 + …,
if a > 0, then the parabola opens upward.
if a < 0, then the parabola opens downward.

Exercise A. 1. Practice drawing the following parabolas with
paper and pencil. Visualize them as the paths of thrown
objects. Make sure pay attention to the symmetry.

Exercise B. Graph the parabolas by taking a table around the
vertex that reflect the symmetry. Find the x and y intercepts.
2. y = –x2
3. y = x2
4. y = x2 – 4
5. y = –x2 + 4
6. y = x2 + 4
7. y = –x2 – 4
8. y = x2 – 2x – 3
9. y = –x2 + 2x + 3
10. y = x2 + 2x – 3
11. y = –x2 – 2x + 3
12. y = x2 – 2x – 8
13. y = –x2 + 2x + 8
14. y = x2 + 2x – 8
15. y = –x2 – 2x + 8
16. a. y = x2 – 4x – 5
b. y = –x2 + 4x + 5
17. a. y = x2 + 4x – 5
b. y = –x2 – 4x + 5
19. y = x2 + 4x – 21
20. y = x2 – 4x – 45
21. y = x2 – 6x – 27
22. y = – x2 – 6x + 27

Exercise C. Graph the following parabolas by plotting the
vertex point, the y–intercept and its reflection.
Find the x intercepts.
24. y = –x2 + 2x + 3
23. y = x2 – 2x – 3
26. y = –x2 – 2x + 3
25. y = x2 + 2x – 3
28. y = –x2 + 2x + 8
27. y = x2 – 2x – 8
30. y = –x2 – 2x + 8
29. y = x2 + 2x – 8
32. y = x2 – 4x – 45
31. y = x2 + 4x – 21
34. y = – x2 – 6x + 27
33. y = x2 – 6x – 27
Exercise D. Graph the following parabolas by plotting the vertex
point, the y–intercept and its reflection. Verify that there is no x
intercepts (i.e. they have complex roots).
35. y = x2 – 2x + 8
36. y = –x2 + 2x – 5
37. y = x2 + 2x + 3
38. y = –x2 – 3x – 4
39. y = 2x2 + 3x + 4
40. y = x2 – 4x + 32

3. y = x2
5. y = –x2 + 4
7. y = –x2 – 4
(0, 4)
(0, –4)
(2, 0)
(–2, 0)
(0,0)
15. y = –x2 – 2x + 8
17. a. y = x2 + 4x – 5
b. y = –x2 – 4x + 5
(–2 , 9)
(–1, 9)
(1, 0)
(–5, 0)
(0, 5)
(1, 0)
(–5, 0)
(2, 0)
(–4 , 0)
(0, –5)
(–2 , –9)
19. y = x2 + 4x – 21
21. y = x2 – 6x – 27
23. y = x2 – 2x – 3
(3, 0)
(–1, 0)
(3, 0)
(–7, 0)
(9, 0)
(–3, 0)
(0, –3)
(0, –21)
(0 , –27)
(0, –21)
(–2 , –25)
(3 , –36)
(–1 , –4)

5 4 the graphs of quadratic equations

by math123b on Aug 28, 2011

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5 4 the graphs of quadratic equations — Presentation Transcript