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By McGraw-Hill Professional
Updated on Oct 4, 2011

The graph of every quadratic function, f ( x ) = ax 2 + bx + c , is a transformation of the graph of y = x 2 . (See Figure 6.1)

Fig. 6.1

When the function is written in the form f ( x ) = a ( xh ) 2 + k , we have a pretty good idea of what its graph looks like: h will cause the graph to shift horizontally, and k will cause it to shift vertically. The point (0, 0) on y = x 2 has shifted to ( h, k ). This point is the vertex . On a parabola that opens up (when a is positive), the vertex is the lowest point on the graph. The vertex is the highest point on a parabola that opens down (when a is negative).

We need to know the vertex when sketching a parabola. Once we have the vertex, we will find two points to its left and two points to its right. We should choose points in such a way that shows the curvature around the vertex and how fast the ends are going up or down. It does not matter which points we choose, but a good rule of thumb is to find h − 2 a, ha, h + a , and h + 2 a . Because a parabola is symmetric about the line x = h (the vertical line that goes through the vertex), the y -values for ha and h + a will be the same and the y -values for h − 2 a and h + 2 a will be the same, too. We will also find the intercepts.

#### Examples

Sketch the graph for the following quadratic functions. Find the y -intercept and the x -intercepts, if any.

• f ( x ) = ( x − 1) 2 − 4

a = 1, h = 1, k = −4 The parabola opens up and the vertex is (1, −4). For the y -intercept, let x = 0 in the function. The y -intercept is (0−1) 2 −4 = 3. For the x -intercepts, let y = 0 and solve for x .

( x − 1) 2 − 4 = 0

( x − 1) 2 = 4 Take the square root of each side.

x − 1 = ±2

x = 1 ± 2 = 1 + 2, 1−2 = 3, −1

The x -intercepts are 3 and − 1.

Table 6.1

 x y Plot this point h − 2 a 1 − 2(1) = −1 (−1 − 1) 2 − 4 = 0 (−1, 0) h − a 1 − 1 = 0 (0 − 1) 2 − 4 = −3 (0,−3) h 1 −4 (1,−4) h + a 1 + 1 = 2 (2 − 1) 2 − 4 = −3 (2,−3) h + 2 a 1 + 2(1) = 3 (3 − 1) 2 − 4 = 0 (3, 0)

Fig. 6.2

• g ( x ) = −2( x + 1) 2 + 18

a = −2, h = −1, k = 18 The parabola opens down, and the vertex is (−1, 18).

 y = −2(0 + 1) 2 + 18 y = 16 −2( x + 1) 2 + 18 = 0 y = 16 − 2( x + 1) 2 = −18 ( x + 1) 2 = 9 x + 1 = ±3 x = −1 ±3 = −1 − 3, −1 + 3 = −4, 2

The y -intercept is 16 and the x -intercepts are −4 and 2.

Table 6.2

 x y Plot this point h − 2 a −1 − 2(−2) = 3 −2(3 + 1) 2 + 18 = −14 (3,−14) h − a −1 − (−2) = 1 −2(1 + 1) 2 + 18 = 10 (1, 10) h −1 18 (−1, 18) h + a −1 + (−2) = −3 −2(−3 + 1) 2 + 18 = 10 (−3, 10) h + 2 a −1 + 2(−2) = −5 −2(−5 + 1) 2 + 18 = −14 (−5,−14)

Fig. 6.3

• The parabola opens up, and the vertex is (−1, 2). Because the parabola opens up ( is positive) and the vertex is above the x -axis ( k = 2 is positive), there will be no x -intercept. If we were to solve the equation , we would not get a real number solution. The y -intercept is .

Table 6.3

 x y Plot this point h − 2 a −1 − 2( ) = −2 (−2 + 1) 2 + 2 = 2 (−2, 2 ) h − a −1 − = −1 (−1 + 1) 2 + 2 = 2 (−1 , 2 ) h −1 2 (−1, 2) h + a −1 + = − (− + 1) 2 + 2 = 2 (− , 2 ) h + 2 a −1 + 2( ) = 0 (0 + 1) 2 + 2 = 2 (0, 2 )

Fig. 6.4

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