Chapter 9.1, Problem 75PFA

To determine

To evaluate which of the given functions has a range of $\left\{y\right|y\le 4\}$

Answer to Problem 75PFA

B $y=-x^2+8x-12$

Explanation of Solution

Given information :

Range of quadratic equation $\left\{y\right|y\le 4\}$

Options.

A $y=x^2-8x+20$

B $y=-x^2+8x-12$

C $y=x^2-8x+12$

D $y=-x^2+8x-20$

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line. The domain of a quadratic equation is all real numbers.

Substituting $x=0$ in the given options.

A $y=0^2-8\times 0+20$

B $y=-0^2+8\times 0-12$

C $y=0^2-8\times 0+12$

D $y=-0^2+8\times 0-20$

For options A and C the value greater than 4, hence they can be eliminated.

Both B and D would have a maximum value as $a<0$ ,

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

For both B and D axis of symmetry is.

  $x=-\frac{8}{2\times (-1)}$

  $x=4$

Therefore, maximum of both quadratic equations is at $x=4$ .

Substituting $x=4$ in the options B and D.

B $y=-4^2+8\times 4-12$

Simplifying

B $y=4$ .

For option D

D $y=-4^2+8\times 4-20$

Simplifying

D $y=-4$ .

The quadratic equation in option D has a maximum value of $y=-4$

Therefore, option B is the correct answer.