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 26E

a.

To determine

To Express: The quadratic function in standard form.

Expert Solution

Answer to Problem 26E

the quadratic function is expressed in standard form as  f(x)=5(x+3)241 

Explanation of Solution

Given: The function is f(x)=(5x2+30x+4)

Calculation:

The quadratic function f(x)=(5x2+30x+4) 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 upwards if a>0 .

Solve the function:

  f(x)=(5x2+30x+4) f(x)=5(x2+6x)+4                     [factor 5 the x terms] f(x)=5(x2+6x+9)+4(59)   [complete the square: add 9 to parentheses, subtract (59) outside] f(x)=5(x+3)241                      [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)=5(x+3)241 

b.

To determine

To Sketch: The graph of the quadratic function.

Expert Solution

Explanation of Solution

Given: The function is f(x)=(5x2+30x+4)

Graph:

The standard form of the function is:

   f(x)=5(x+3)241 

From the standard form it is observed that the graph is a parabola that opens upward and has vertex (3,41) . As an aid to sketching the graph, find the intercepts.

The y-intercept=f(0)=4 and The x-intercept is 3±2055 and 0 . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: f(x)=(5x2+30x+4)

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 26E

c.

To determine

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

Expert Solution

Answer to Problem 26E

The value of minima is f(3)=(41)

Explanation of Solution

Given: The function is f(x)=(5x2+30x+4)

Calculation:

From the above graph it is seen that the parabola opens upward, since the coefficient of x2 is positive, f has minimum value. The value of minima is f(3)=(41)

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