BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 3.1, Problem 28E

a.

To determine

To Express: The quadratic function in standard form.

Expert Solution

Answer to Problem 28E

the quadratic function is expressed in standard form as f(x)=1(x2+3)2+10

Explanation of Solution

Given: The function is f(x)=(16xx2)

Calculation:

The quadratic function f(x)=(16xx2) is expressed in standard form as:

  f(x)=a(xh)2+k , by completing the square. The graph of the function f is a parabola with vertex (h,k)

The parabola opens downwards if a<0 .

Solve the function:

  f(x)=(16xx2)f(x)=(x26x+1)f(x)=1(x2+6x)+1                       [factor the x terms]f(x)=1(x2+6x+9)+1(19)   [complete the square: add 9 to the parentheses, subtract (19) outside]f(x)=1(x2+3)2+10                      [factor and multiply]

On comparing the above equation with standard form f(x)=a(xh)2+k ,

Therefore, the quadratic function is expressed in standard form as f(x)=1(x2+3)2+10

b.

To determine

To Sketch: The graph of the quadratic function.

Expert Solution

Explanation of Solution

Given: The function is f(x)=(16xx2)

Graph:

The standard form of the function is:

  f(x)=1(x2+3)2+10

From the standard form it is observed that the graph is a parabola that opens upward and has vertex (3,10) . As an aid to sketching the graph, find the intercepts.

The y-intercept=f(0)=1 and The x-intercept is 3±10 and 0 . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: f(x)=(16xx2)

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 28E

c.

To determine

To Find: The maximum or minimum value of the function.

Expert Solution

Answer to Problem 28E

The value of maxima is f(3)=(10)

Explanation of Solution

Given: The function is f(x)=(16xx2)

Calculation:

From the above graph it is seen that the parabola opens downward, since the coefficient of x2 is negative, f has maximum value. The value of maxima is f(3)=(10)

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