Quadratic Functions
Graphs of Quadratic Functions
The graph of any quadratic function is called a parabola.
Parabolas are shaped like cups, as shown in the graph below. If
the coefficient of x2 is positive, the parabola opens upward;
otherwise, the parabola opens downward. The vertex (or turning
point) is the minimum or maximum point.
The Standard Form of a
Quadratic Function
The quadratic function
f (x) = a(x - h)2 + k,
a0
is in standard form. The graph of f is a parabola
whose vertex is the point (h, k). The parabola is
symmetric to the line x = h. If a > 0, the parabola
opens upward; if a < 0, the parabola opens
downward.
Graphing Parabolas With
Equations in Standard Form
To graph f (x) = a(x - h)2 + k:
1. Determine whether the parabola opens upward or
downward. If a > 0, it opens upward. If a < 0, it
opens downward.
2. Determine the vertex of the parabola. The vertex is
(h, k).
3. Find any x-intercepts by replacing f (x) with 0.
Solve the resulting quadratic equation for x.
4. Find the y-intercept by replacing x with zero.
5. Plot the intercepts and vertex. Connect these
points with a smooth curve that is shaped like a
cup.
Text Example
Graph the quadratic function f (x) = -2(x - 3)2 +
8.
Solution We can graph this function by following the steps in the preceding
box. We begin by identifying values for a, h, and k.
Standard form
f (x) = a(x - h)2 + k
a = -2
h=3
k=8
Given equation f (x) = -2(x - 3)2 + 8
Step 1 Determine how the parabola opens. Note that a, the coefficient of
x 2, is -2. Thus, a < 0; this negative value tells us that the parabola opens
downward.
Text Example cont.
Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h =
3 and k = 8, the parabola has its vertex at (3, 8).
Step 3
Find the x-intercepts. Replace f(x) with 0 in f(x) = -2(x - 3)2 + 8.
0 = -2(x - 3)2 + 8
Solve for x. Add 2(x - 3)2 to both sides of
the equation.
2(x - 3) = 8
2
(x - 3)2 = 4
(x - 3) = 2
x - 3 = -2
x=1
Find x-intercepts, setting f (x) equal to zero.
or
or
Divide both sides by 2.
Apply the square root method.
x-3=2
x=5
Express as two separate equations.
Add 3 to both sides in each equation.
The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).
Text Example cont.
Step 4
Find the y-intercept. Replace x with 0 in f(x) = -2(x - 3)2 + 8.
f(0) = -2(0 - 3)2 + 8 = -2(-3)2 + 8 = -2(9) + 8 = -10
The y-intercept is –10. The parabola passes through (0, -10).
Step 5 Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5,
and a y-intercept at –10, the axis of symmetry is the vertical line whose
equation is x = 3.
The Vertex of a Parabola Whose
Equation Is f (x) = ax 2 + bx + c
Consider the parabola defined by the
quadratic function f (x) = ax 2 + bx + c.
The parabola's vertex is at
 -b  -b  
 , f  
 2a  2a  
Example
Graph the quadratic function f (x) = -x + 6x -.
2
Solution:
Step 1 Determine how the parabola opens. Note that a, the
coefficient of x 2, is -1. Thus, a < 0; this negative value tells us that
the parabola opens downward.
Step 2 Find the vertex. We know the x-coordinate of the vertex is –
b/2a.
We identify a, b, and c to substitute the values into the equation for the
x-coordinate:
x = -b/(2a) = -6/2(-1) = 3.
The x-coordinate of the vertex is 3. We substitute 3 for x in the
equation of the function to find2 the y-coordinate:
y = f (3) = -3 + 6(3) - 2 = -9 + 18 - 2 = 7
the parabola has its vertex at (3,7).
Example
Graph the quadratic function f (x) = -x2 + 6x -.
Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x
- 2. 0 = -x2 + 6x - 2
a = -1,b = 6,c = -2
-b  b 2 - 4ac
x=
2a
2
-6  6 - 4(-1)(-2)
=
2(-1)
-6  36 - 8
-2
-6  28 -6  2 7
=
=
-2
-2
= 3 7
=
Example
Graph the quadratic function f (x) = -x2 + 6x -.
Step 4
- 2.
Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x
f (0) = -02 + 6 • 0 - 2 = -
The y-intercept is –2. The parabola passes 10through (0, -2).
8
Step 5 Graph the parabola.
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
Minimum and Maximum:
Quadratic Functions
•
Consider f(x) = ax2 + bx +c.
1. If a > 0, then f has a minimum that occurs at
x = -b/(2a). This minimum value is f(-b/(2a)).
2. If a < 0, the f has a maximum that occurs at
x = -b/(2a). This maximum value is f(-b/(2a)).
Strategy for Solving Problems Involving
Maximizing or Minimizing Quadratic
Functions
1.
2.
3.
4.
5.
Read the problem carefully and decide which quantity
is to be maximized or minimized.
Use the conditions of the problem to express the
quantity as a function in one variable.
Rewrite the function in the form f(x) = ax2 + bx +c.
Calculate -b/(2a). If a > 0, then f has a minimum that
occurs at x = -b/(2a). This minimum value is f(-b/(2a)).
If a < 0, the f has a maximum that occurs at x = b/(2a). This maximum value is f(-b/(2a)).
Answer the question posed in the problem.