Question

Find the vertex, zero(s), and y-intercept of the graph of y = –x2 + 4x + 32.

Answer 1

Answer:

Option (a) is correct.

Vertex: (2,36); zeros: (–4,0), (8,0) y-intercept: (0,32)

Step-by-step explanation:

Consider the given equation $y=-x^2+4x+32$

We have to find vertex, zero(s), and y-intercept.

First we find the vertex, For The general form of a quadratic is $y=ax^2+bx+c$

the coordinate of the vertex (h, k)  is given as $h=\frac{-b}{2a}$ and

$k=\frac{4ac-b^2}{4a}$

Here, a= -1 , b= 4 and  c = 32

$h=\frac{-b}{2a}=\frac{-4}{-2}=2$ and,

$k=\frac{4ac-b^2}{4a}=\frac{-128-16}{-4}=36$

Thus vertex of $y=-x^2+4x+32$  is ( 2, 36)

Now, we find the zeros,

Put y = 0, we get,

$y=-x^2+4x+32=0$

This is a quadratic equation of the form $ax^2+bx+c=0$

Hence, we can find zero using middle term splitting method,

4x can be written as 8x - 4x

Thus, $-x^2+4x+32=0$

$\Rightarrow -x^2+8x-4x+32=0$

$\Rightarrow x(-x+8)+4(-x+8)=0$

$\Rightarrow (-x+8)(x+4)=0$

$\Rightarrow (-x+8)=0$ or $\Rightarrow (x+4)=0$

$\Rightarrow x=8$ or $\Rightarrow x=-4$

Thus, zeros are (-4,0) and (8,0) .

Now to calculate y intercept put x = 0 in $y=-x^2+4x+32$

We get , y= 32.

The same can be seen through graph as below.

Thus, option (a) is correct.

Answer 2

Answer:

The correct option is A.

Step-by-step explanation:

The given equation is

$y=-x^2+4x+32$

Put x=0, in the given equation.

$y=-(0)^2+4(0)+32=32$

The y-intercept is (0,32).

Put y=0, to find the x-intercept.

$0=-x^2+4x+32$

$0=-x^2+8x-4x+32$

$0=-x(x-8)-4(x-8)$

$0=-(x+4)(x-8)$

$x=-4,8$

Therefore the y-intercepts are (-4,0) and (8,0).

The vertex of a parabola $f(x)=ax^2+bx+c$ is

$(-\frac{b}{2a},f(-\frac{b}{2a}))$

$-\frac{b}{2a}=-\frac{4}{2(-1)}=2$

Put x=2 in the given function.

$y=-(2)^2+4(2)+32$

$y=-4+8+32$

$y=36$

The vertex is (2,36).

Therefore option A is correct.