1. Graphing Quadratic Functions
Digital Lesson
Graphing Quadratic Functions

2. Graphing Quadratic Functions Using a Table
Graph f(x) = x2 – 4x + 3 by using a table.
Make a table. Plot enough ordered pairs to see both sides of the curve. x
f(x)= x2 – 4x + 3
(x, f(x))
f(0)= (0)2 – 4(0) + 3
(0, 3) 1
f(1)= (1)2 – 4(1) + 3
(1, 0) 2
f(2)= (2)2 – 4(2) + 3
(2,–1) 3
f(3)= (3)2 – 4(3) + 3
(3, 0) 4
f(4)= (4)2 – 4(4) + 3
(4, 3)

3. Example 1 Continued
f(x) = x2 – 4x + 3 • • • • •

5. Example 1: Translating Quadratic Functions
Use the graph of f(x) = x2 as a guide, describe the transformations and then graph each function.
g(x) = (x – 2)2 + 4
Identify h and k.
g(x) = (x – 2)2 + 4 h k
Because h = 2, the graph is translated 2 units right. Because k = 4, the graph is translated 4 units up. Therefore, g is f translated 2 units right and 4 units up.

6. Example 2
Use the graph of f(x) = x2 as a guide, describe the transformations and then graph each function.
g(x) = (x + 2)2 – 3
Identify h and k.
g(x) = (x – (–2))2 + (–3) h k
Because h = –2, the graph is translated 2 units left. Because k = –3, the graph is translated 3 units down. Therefore, g is f translated 2 units left and 4 units down.

7. Example 3
Using the graph of f(x) = x2 as a guide, describe the transformations and then graph each function.
g(x) = x2 – 5
Identify h and k.
g(x) = x2 – 5 k
Because h = 0, the graph is not translated horizontally. Because k = –5, the graph is translated 5 units down. Therefore, g is f is translated 5 units down.

9. Example 3A: Reflecting, Stretching, and Compressing Quadratic Functions
Using the graph of f(x) = x2 as a guide, describe the transformations and then graph each function. 1 ( )
g x =- x 2 4
Because a is negative, g is a reflection of f across the x-axis.
Because |a| = , g is a vertical compression of f by a factor of .

10. If a parabola opens upward, it has a lowest point
If a parabola opens upward, it has a lowest point. If a parabola opens downward, it has a highest point. This lowest or highest point is the vertex of the parabola.
The parent function f(x) = x2 has its vertex at the origin. You can identify the vertex of other quadratic functions by analyzing the function in vertex form. The vertex form of a quadratic function is f(x) = a(x – h)2 + k, where a, h, and k are constants.

11. Because the vertex is translated h horizontal units and k vertical from the origin, the vertex of the parabola is at (h, k).
When the quadratic parent function f(x) = x2 is written in vertex form, y = a(x – h)2 + k, a = 1, h = 0, and k = 0.
Helpful Hint

12. Example 4: Writing Transformed Quadratic Functions
Use the description to write the quadratic function in vertex form.
The parent function f(x) = x2 is vertically stretched by a factor of and then translated 2 units left and 5 units down to create g.
Step 1 Identify how each transformation affects the constant in vertex form.
Vertical stretch by : 4 3 a =
Translation 2 units left: h = –2
Translation 5 units down: k = –5

13. Example 4: Writing Transformed Quadratic Functions
Step 2 Write the transformed function.
g(x) = a(x – h)2 + k
Vertex form of a quadratic function
= (x – (–2))2 + (–5)
Substitute for a, –2 for h, and –5 for k.
= (x + 2)2 – 5
Simplify.
g(x) = (x + 2)2 – 5

14. Check Graph both functions on a graphing calculator
Check Graph both functions on a graphing calculator. Enter f as Y1, and g as Y2. The graph indicates the identified transformations. f g

15. Check It Out! Example 4a
Use the description to write the quadratic function in vertex form.
The parent function f(x) = x2 is vertically compressed by a factor of and then translated units right and 4 units down to create g.
Step 1 Identify how each transformation affects the constant in vertex form.
Vertical compression by : a =
Translation 2 units right: h = 2
Translation 4 units down: k = –4

16. Check It Out! Example 4a Continued
Step 2 Write the transformed function.
g(x) = a(x – h)2 + k
Vertex form of a quadratic function
= (x – 2)2 + (–4)
Substitute for a, 2 for h, and –4 for k.
= (x – 2)2 – 4
Simplify.
g(x) = (x – 2)2 – 4

17. Check It Out! Example 4a Continued
Check Graph both functions on a graphing calculator. Enter f as Y1, and g as Y2. The graph indicates the identified transformations. f g

18. Example
The parent function f(x) = x2 is vertically stretched by a factor of 3 and translated 4 units right and 2 units up to create g. Write g in vertex form.
g(x) = 3(x – 4)2 + 2

19. is called a quadratic function.
Let a, b, and c be real numbers a  0. The function f (x) = ax2 + bx + c
is called a quadratic function.
The graph of a quadratic function is a parabola.
Every parabola is symmetrical about a line called the axis (of symmetry). x y
The intersection point of the parabola and the axis is called the vertex of the parabola.
f (x) = ax2 + bx + c
vertex
axis
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Quadratic function

20. The leading coefficient of ax2 + bx + c is a.y
a > 0 opens upward
When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum.
f(x) = ax2 + bx + c
vertex minimum x y
vertex maximum
When the leading
coefficient is negative, the parabola opens downward and the vertex is a maximum.
f(x) = ax2 + bx + c
a < 0 opens downward
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Leading Coefficient

21. Simple Quadratic Functions
The simplest quadratic functions are of the form
f (x) = ax2 (a  0)
These are most easily graphed by comparing them with the graph of y = x2.
Example: Compare the graphs of
, and 5 y x -5
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Simple Quadratic Functions

22. Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.
f (x) = (x – 3)2 + 2 is the same shape as the graph of
g (x) = (x – 3)2 shifted upwards two units.
g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units.
- 4 x y 4
f (x) = (x – 3)2 + 2
g (x) = (x – 3)2
y = x 2
vertex
(3, 2)
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Example: f(x) = (x –3)2 + 2

23. The vertex of the graph of f (x) = ax2 + bx + c (a  0)
Vertex of a Parabola
The vertex of the graph of f (x) = ax2 + bx + c (a  0)
Example: Find the vertex of the graph of f (x) = x2 – 10x + 22.
f (x) = x2 – 10x original equation
a = 1, b = –10, c = 22
At the vertex,
So, the vertex is (5, -3).
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Vertex of a Parabola

24. Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is:
The path is a parabola opening downward. The maximum height occurs at the vertex.
At the vertex,
So, the vertex is (9, 15).
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Example: Basketball

25. Let x represent the width of the corral and 120 – 2x the length.
Example: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area?
barn
corral x
120 – 2x
Let x represent the width of the corral and 120 – 2x the length.
Area = A(x) = (120 – 2x) x = –2x x
The graph is a parabola and opens downward.
The maximum occurs at the vertex where
a = –2 and b = 120
120 – 2x = 120 – 2(30) = 60
The maximum area occurs when the width is 30 feet and the length is 60 feet.
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Example: Maximum Area

26. Quadratic Function in Standard Form
The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0)
The graph is a parabola opening upward if a  0 and opening downward if a  0. The axis is x = h, and the vertex is (h, k).
Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis and vertex. x y
f (x) = 2x2 + 4x – 1
x = –1
f (x) = 2x2 + 4x – original equation
f (x) = 2( x2 + 2x) – factor out 2
f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square
f (x) = 2( x + 1)2 – standard form
a > 0  parabola opens upward like y = 2x2.
(–1, –3)
h = –1, k = –3  axis x = –1, vertex (–1, –3).
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Quadratic Function in Standard Form

27. Vertex and x-Intercepts
Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7. x y 4
f (x) = – x2 + 6x original equation
(3, 16)
x = 3
f (x) = – ( x2 – 6x) factor out –1
f (x) = – ( x2 – 6x + 9) complete the square
f (x) = – ( x – 3) standard form
a < 0  parabola opens downward.
h = 3, k = 16  axis x = 3, vertex (3, 16).
Find the x-intercepts by solving –x2 + 6x + 7 = 0.
(7, 0)
(–1, 0)
(–x + 7 )( x + 1) = factor
x = 7, x = –1 x-intercepts (7, 0), (–1, 0)
f(x) = –x2 + 6x + 7
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Vertex and x-Intercepts

28. Vertex Form of a Quadratic
f(x) = a(x – h)2 + k
Where (h, k) is the vertex.
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29. f (x) = a(x – h)2 + k standard form
Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1). y x
y = f(x)
(0, 1)
(2, –1)
f (x) = a(x – h)2 + k standard form
f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k)
Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1
1 = 4a –1 and
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Example: Parabola