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 3E
To determine

To fill: The statement “The graph of f(x)=3(x2)26 is a parabola that opens _____________ with vertex at (___,___) and f(2)=___ is the (minimum/maximum)_____________ value of f ”.

Expert Solution

Answer to Problem 3E

The graph of f(x)=3(x2)26 is a parabola that opens upward with vertex at (2,6) and f(2)=6 is the minimum value of f.

Explanation of Solution

Definition used:

The standard form of the quadratic function is f(x)=a(xh)2+k . The graph of this function turns out to be a parabola with vertex set (h,k) . The parabola opens upward if a>0 and opens downward if a<0 .

The maximum or minimum value of f occurs at x=h . If a<0 , then the maximum value of f is f(h)=k and if a>0 , then the minimum value of f is f(h)=k .

Calculation:

Compare the standard form of quadratic equation f(x)=a(xh)2+k with f(x)=3(x2)26 and notice that a=3 , h=2 and k=6 .

From the definition used above, a=3 which is greater than zero and therefore the parabola opens upward.

Notice that the vertex is at (h,k) which is (2,6) .

Since a>0 , the minimum value of f is at f(h)=k is,

f(h)=kf(2)=6

Therefore, at f(2)=6 the function has a minimum value.

Hence, the graph of f(x)=3(x2)26 is a parabola that opens upward with vertex at (2,6) and f(2)=6 is the minimum value of f.

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