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 4E
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 4E

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

Explanation of Solution

Formula used:

Standard form of a quadratic function:

The standard form of a quadratic function f(x)=ax2+bx+c is f(x)=a(xh)2+k, where (h,k) is the vertex of the parabola. And the parabola opens upward if a>0 or 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.

Then, a=3, h=2 and k=6.

From the definition used above, a=3 which is less than zero and therefore the parabola of the function opens downward.

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

Since a<0, the maximum value of f is at f(h)=k is,

f(h)=kf(2)=6

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

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

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