Answer:
The correct option is A.
Step-by-step explanation:
The given equation is
[tex]y=-x^2+4x+32[/tex]
Put x=0, in the given equation.
[tex]y=-(0)^2+4(0)+32=32[/tex]
The y-intercept is (0,32).
Put y=0, to find the x-intercept.
[tex]0=-x^2+4x+32[/tex]
[tex]0=-x^2+8x-4x+32[/tex]
[tex]0=-x(x-8)-4(x-8)[/tex]
[tex]0=-(x+4)(x-8)[/tex]
[tex]x=-4,8[/tex]
Therefore the y-intercepts are (-4,0) and (8,0).
The vertex of a parabola [tex]f(x)=ax^2+bx+c[/tex] is
[tex](-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]
[tex]-\frac{b}{2a}=-\frac{4}{2(-1)}=2[/tex]
Put x=2 in the given function.
[tex]y=-(2)^2+4(2)+32[/tex]
[tex]y=-4+8+32[/tex]
[tex]y=36[/tex]
The vertex is (2,36).
Therefore option A is correct.
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 [tex]y=-x^2+4x+32[/tex]
We have to find vertex, zero(s), and y-intercept.
First we find the vertex, For The general form of a quadratic is [tex]y=ax^2+bx+c[/tex]
the coordinate of the vertex (h, k) is given as [tex]h=\frac{-b}{2a}[/tex] and
[tex]k=\frac{4ac-b^2}{4a}[/tex]
Here, a= -1 , b= 4 and c = 32
[tex]h=\frac{-b}{2a}=\frac{-4}{-2}=2[/tex] and,
[tex]k=\frac{4ac-b^2}{4a}=\frac{-128-16}{-4}=36[/tex]
Thus vertex of [tex]y=-x^2+4x+32[/tex] is ( 2, 36)
Now, we find the zeros,
Put y = 0, we get,
[tex]y=-x^2+4x+32=0[/tex]
This is a quadratic equation of the form [tex]ax^2+bx+c=0[/tex]
Hence, we can find zero using middle term splitting method,
4x can be written as 8x - 4x
Thus, [tex]-x^2+4x+32=0[/tex]
[tex]\Rightarrow -x^2+8x-4x+32=0[/tex]
[tex]\Rightarrow x(-x+8)+4(-x+8)=0[/tex]
[tex]\Rightarrow (-x+8)(x+4)=0[/tex]
[tex]\Rightarrow (-x+8)=0[/tex] or [tex]\Rightarrow (x+4)=0[/tex]
[tex]\Rightarrow x=8[/tex] or [tex]\Rightarrow x=-4[/tex]
Thus, zeros are (-4,0) and (8,0) .
Now to calculate y intercept put x = 0 in [tex]y=-x^2+4x+32[/tex]
We get , y= 32.
The same can be seen through graph as below.
Thus, option (a) is correct.
