Question

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

A. Vertex: (2,36); zeros: (–4,0), (8,0) y-intercept: (0,32)
B. Vertex: (36,2); zeros: (–4,0), (8,0) y-intercept: (32,0)
C. Vertex: (2,3); zeros: (–4,0), (8,0) y-intercept: (0,16)
D. Vertex: (2,36); zeros: (–8,0), (4,0) y-intercept: (0,32)

Answer 1

Option (a) is correct.

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

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=(-b)/(2a)` and

`k=(4ac-b^2)/(4a)`

Here, a= -1 , b= 4 and c = 32

`h=(-b)/(2a)=(-4)/(-2)=2` and,

`k=(4ac-b^2)/(4a)=(-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

The correct option is A.

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

`(-(b)/(2a),f(-(b)/(2a)))`

`-(b)/(2a)=-(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.