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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 x 2 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:

2. De termine whether the parabola opens upward or
downward. If a > 0, it opens upward. If a < 0, it opens
downward.

3. Determine the vertex of the parabola. The vertex is (h,
k).

4. Find any x-intercepts
by replacing f (x) with 0. Solve
the resulting quadratic equation for x .

5. Find the y-intercept
by replacing x with zero .

6. 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 fol lowing the steps in the preceding
box. We begin by identifying values for a , h, and k.

Standard form
Given equation

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.

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 Find x-intercepts,
setting f (x) equal to zero.
2(x - 3)2 = 8 Solve for x . Add 2(x - 3)2 to both sides of
the equation.
(x - 3)2 = 4 Divide both sides by 2.
(x - 3) = ±2 Apply the square root method .
x - 3 = -2 or x - 3 = 2 Express as two separate equations.
x = 1 or x = 5 Add 3 to both sides in each equation.

The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).

Step 4 Find the y-intercept. Replace x with 0 in f (x) = -2(x - 3)2 + 8.

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) = ax2 + bx + c

• Consider the parabola defined by the
quadratic function

• f (x) = ax2 + bx + c. The parabola's vertex is
at

Example

Graph the quadratic function f (x) = -x2 + 6x -2.

Solution:

Step 1 Determine how the parabola opens. Note that a, the
coefficient of x2, 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 x =
-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 find the y-coordinate:
y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, the parabola has its vertex at (3,7).

• Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x
- 2. a = -1,b = 6,c = -2

• 0 = -x2 + 6x - 2


• Step 4 Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x -
2.

The y-intercept is –2. The parabola passes through (0, -2).

Step 5 Graph the parabola.

Minimum and Maximum:
Quadratic Functions


• Consider f(x) = ax2 + bx +c.

2. If a > 0, then f has a minimum that occurs at
x = -b/(2a). This minimum value is f(-b/(2a)).

3. If a < 0, the f has a maximum that occurs at
x = -b/(2a). This maximum value is f (-b/(2a)).

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