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# Quadratic Functions Help (page 3)

based on 4 ratings
By McGraw-Hill Professional
Updated on Oct 4, 2011

#### Practice

For Problems 1-3, sketch the graph and identify the vertex and intercepts.

1. y = −( x − 1) 2 + 4

1. Rewrite in the form f ( x ) = a ( xh ) 2 + k , using completing the square.
2. Find the maximum or minimum functional value for g ( x ) = −0.002 x 2 + 5 x + 150.
3. &Find an equation for the quadratic function whose vertex is (2, 5) and whose graph contains the point (−8, 15).

#### Solutions

1. The vertex is (1,4). The y -intercept is −(0 − 1) 2 + 4 = 3.

−( x − 1) 2 + 4 = 0

−( x − 1) 2 = −4

( x − 1) 2 = 4

( x − 1) = ±2

x = 1 ± 2 = 1 + 2, 1 − 2 = 3, − 1

The x -intercepts are 3 and − 1.

Fig. 6.5

2. The vertex is (−1, 2). The y -intercept is . There are two ways we can tell that there are no x -intercepts. The parabola opens up and the vertex is above the x -axis, so the parabola is always above the x -axis. Also, the equation has no real number solution.

Fig. 6.6

3. The vertex is .

The y -intercept is .

The x -intercepts are −6 and 4.

Fig. 6.7

4.

5. This function has a maximum value because a = −0.002 is negative. The answer is k .

The maximum functional value is 3275.

6. h = 2, k = 5 which makes y = a ( xh ) 2 + k become y = a ( x − 2) 2 + 5. We can find a by letting x = − 8 and y = 15.

y = a ( x − 2) 2 + 5

15 = a (−8−2) 2 + 5

10 = a (−10) 2

10 = 100 a

0.1 = a

The equation is y = 0.1 ( x − 2) 2 + 5.

Practice problems for this concept can be found at: Quadratic Functions Practice Test.

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