Answer:
The correct option is C) function (x-5)(x+1) has vertex (2,-9).
Step-by-step explanation:
The vertex of an up down facing parabola of the form y=ax²+bx+c is [tex]x_v=-\frac{b}{2a}[/tex]
option A) -(x-3)²
Rewrite [tex]y=-\left(x-3\right)^2[/tex] in the form [tex]y=ax^{2}+bx+c[/tex]
Expand [tex]-\left(x-3\right)^{2}[/tex]
[tex]\left(x^{2}-6x+9\right)[/tex]
The parabola parameters are: a = - 1, b = 6, c = - 9
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{6}{2\left(-1\right)}[/tex]
simplify, 3
Plugin [tex]x_v = 3[/tex] to find the [tex]y_v[/tex] value
[tex]y_v=-3^{2}+6\times 3 -9[/tex]
[tex]y_v=-0[/tex]
If a<0, then the vertex is a maximum value.
If a>0, then the vertex is a minimum value.
since, a = - 1
Maximum (3,0)
option B) (x+8)²
Rewrite [tex]y=\left(x+8\right)^{2}[/tex] in the form [tex]y=ax^{2}+bx+c[/tex]
Expand [tex]\left(x+8\right)^{2}[/tex]
[tex]\left(x^{2}+16x+64\right)[/tex]
The parabola parameters are: a = 1, b = 16, c = 64
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{16}{2\left(1\right)}[/tex]
simplify, - 8
Plugin [tex]x_v = -8 [/tex] to find the [tex]y_v[/tex] value
[tex]y_v=-8^{2}+16(-8)+64[/tex]
[tex]y_v=-0[/tex]
If a<0, then the vertex is a maximum value.
If a>0, then the vertex is a minimum value.
since, a = 1
Minimum (-8,0)
option C) (x-5)(x+1)
Rewrite [tex]y=(x-5)(x+1)[/tex] in the form [tex]y=ax^{2}+bx+c[/tex]
Expand [tex]y=(x-5)(x+1)[/tex]
[tex]\left(x^{2}-4x-5\right)[/tex]
The parabola parameters are: a = 1, b = -4, c = -5
[tex]x_v=-\frac{b}{2a}[/tex]
[tex]x_v=-\frac{-4}{2\left(1\right)}[/tex]
simplify, 2
Plugin [tex]x_v = 2 [/tex] to find the [tex]y_v[/tex] value
[tex]y_v=2^{2}-4(2)-5[/tex]
[tex]y_v=-9[/tex]
If a<0, then the vertex is a maximum value.
If a>0, then the vertex is a minimum value.
since, a = 1
Minimum (2,-9)
Hence, the correct option is C) function (x-5)(x+1) has vertex (2,-9).
