For Problems 1 − 34 , graph each of the following linear and quadratic functions. ( Objectives 1 and 3 ) f ( x ) = − 3 x 2 + 2

Q: Expert Answer

To determine To plot: The graph of the function f ( x ) = − 3 x 2 + 2 . Explanation Procedure: Graph of a quadratic function of the form, f ( x ) = a x 2 + b x + c , is shown by the following steps. 1 . Check whether the parabola opens upward (if a > 0 ) or downward (if a < 0 ). 2 . Calculate − b 2 a , which is the y coordinate of the vertex of parabola. 3 . Calculate f ( − b 2 a ) , which is y coordinate of the vertex of parabola. 4 . Locate the another point on the graph of parabola, and also locate the image across the line of symmetry, x = − b 2 a . Approach: Consider the given function as shown below, f ( x ) = − 3 x 2 + 2 … ( 1 ) Compare the function with the standard form of a parabola f ( x ) = a x 2 + b x + c . Here, a = − 3 , b = 0 and c = 2 . Since the value of a is negative, the parabola will open downward. The x coordinate of the vertex is given by, x = − b 2 a … ( 2 ) Substitute − 3 for a and 0 for b in the equation ( 2 ) to calculate x coordinate of the vertex. x = − ( 0 ) 2 ( 2 ) x = 0 The x coordinate of the vertex is 0 . Now, find the value of f ( − b 2 a ) and substitute 0 for x in equation ( 1 )

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Answer 1

BuyFind

Intermediate Algebra

10th Edition

Jerome E. Kaufmann + 1 other

Publisher: Cengage Learning

ISBN: 9781285195728

Solutions

Chapter

Section

Answer 2

BuyFind

Intermediate Algebra

10th Edition

Jerome E. Kaufmann + 1 other

Publisher: Cengage Learning

ISBN: 9781285195728

Solutions

Chapter

Section

Answer 3

Solutions

Chapter

Section

Answer 4

Chapter

Section

Answer 5

To determine

To plot:

The graph of the function f(x)=−3x2+2.

Explanation of Solution

Procedure:

Graph of a quadratic function of the form, f(x)=ax2+bx+c, is shown by the following steps.

1. Check whether the parabola opens upward (if a>0) or downward (if a<0).

2. Calculate −b2a, which is the y coordinate of the vertex of parabola.

3. Calculate f(−b2a), which is y coordinate of the vertex of parabola.

4. Locate the another point on the graph of parabola, and also locate the image across the line of symmetry, x=−b2a.

Approach:

Consider the given function as shown below,

f(x)=−3x2+2…(1)

Compare the function with the standard form of a parabola f(x)=ax2+bx+c.

Here, a=−3, b=0 and c=2.

Since the value of a is negative, the parabola will open downward.

The x coordinate of the vertex is given by,

x=−b2a…(2)

Substitute −3 for a and 0 for b in the equation (2) to calculate x coordinate of the vertex.

x=−(0)2(2)x=0

The x coordinate of the vertex is 0.

Now, find the value of f(−b2a) and substitute 0 for x in equation (1)

For Problems 1 − 34 , graph each of the following linear and quadratic functions. ( Objectives 1 and 3 ) f ( x ) = − 3 x 2 + 2

Intermediate Algebra

Solutions

Intermediate Algebra