5 3 the graphs of quadratic equations

Graphs of Quadratic Equations
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The graphs of 1st degree equations Ax + By = C
are straight lines hence they are called linear equations.

If we have a linear function in the form of y = f(x) = mx + b,
then we may eyeball the graph, starting from the y-intercept b,

For example, for y = -2x + 3, we start at the point (0, 3).
the graph of y = -2x + 3

eyeball the tilt using the slope m.

The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,

i.e. another point is located 2 down and 1 to the right from (0, 3)

(0, 3)
∆y = -2

(0, 3)
(1, 1)∆y = -2
∆x = 1

(0, 3)
(1, 1)∆y = -2
∆x = 1
and the line contains both points.

and the line contains both points.
The graphs of 2nd degree equations
y = ax2 + bx + c
are not straight.
They are curves.
We will develop a method
for eyeballing the graphs of
y = ax2 + bx + c in this section.
(0, 3)
(1, 1)∆y = -2
∆x = 1

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

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

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

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).

down paths).
Properties of Parabolas:

down paths).
• Parabolas are symmetric with respect to a center line

down paths).
• The highest or lowest point of the parabola sits on the
center line. This point is called the vertex, or the tip.

down paths).
center line. This point is called the vertex, or the tip.
the vertex
the vertex

down paths).
center line. This point is called the vertex.
The vertex is the starting point for graphing a parabola,
i.e. the graph of y = ax2 + bx + c, a 2nd degree function.
the vertex
the vertex

down paths).
center line. This point is called the vertex.
Vertex Formula: The vertex of y = ax2 + bx + c is at x = .
–b
2a
The vertex is the starting point for graphing a parabola,
i.e. the graph of y = ax2 + bx + c, a 2nd degree function.
the vertex
the vertex

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
Note the y values are symmetric
around the vertex just as the
points.
–12
–15
–16
–15
–12

Vertex: set x = = 2
–(–4)
2(1)
x y
0
1
2
3
4
points. (If they are not, check
your calculation.)
–12
–15
–16
–15
–12

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

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

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

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

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

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

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

equation 0 = ax2 + bx + c which may or may not have real
number solutions.

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,

know another point on the parabola, the reflection of the point
across the center is also on the parabola.

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.

1. 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

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
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5
Example C. Graph y = –x2 + 2x + 15 (1, 16)

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

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

-b
2a
–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5
(0, 15) (2, 15)
(-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.
4. y = x2 – 4 5. y = –x2 + 4
2. y = –x2 3. y = x2
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.
23. y = x2 – 2x – 3 24. y = –x2 + 2x + 3
25. y = x2 + 2x – 3 26. y = –x2 – 2x + 3
27. y = x2 – 2x – 8 28. y = –x2 + 2x + 8
29. y = x2 + 2x – 8 30. y = –x2 – 2x + 8
31. y = x2 + 4x – 21 32. y = x2 – 4x – 45
33. y = x2 – 6x – 27 34. 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

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

5 3 the graphs of quadratic equations

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