Question

Which function has a vertex at (2,-9)​

Answer 1

Answer:

C. f(x) = (x – 5)(x + 1)

Step-by-step explanation:

Answer 2

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 $x_v=-\frac{b}{2a}$

option A) -(x-3)²

Rewrite $y=-\left(x-3\right)^2$ in the form $y=ax^{2}+bx+c$

Expand $-\left(x-3\right)^{2}$

$\left(x^{2}-6x+9\right)$

The parabola parameters are: a = - 1, b = 6, c = - 9

$x_v=-\frac{b}{2a}$

$x_v=-\frac{6}{2\left(-1\right)}$

simplify, 3

Plugin $x_v = 3$ to find the $y_v$ value

$y_v=-3^{2}+6\times 3 -9$

$y_v=-0$

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 $y=\left(x+8\right)^{2}$ in the form $y=ax^{2}+bx+c$

Expand $\left(x+8\right)^{2}$

$\left(x^{2}+16x+64\right)$

The parabola parameters are: a = 1, b = 16, c = 64

$x_v=-\frac{b}{2a}$

$x_v=-\frac{16}{2\left(1\right)}$

simplify, - 8

Plugin $x_v = -8$ to find the $y_v$ value

$y_v=-8^{2}+16(-8)+64$

$y_v=-0$

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 $y=(x-5)(x+1)$ in the form $y=ax^{2}+bx+c$

Expand $y=(x-5)(x+1)$

$\left(x^{2}-4x-5\right)$

The parabola parameters are: a = 1, b = -4, c = -5

$x_v=-\frac{b}{2a}$

$x_v=-\frac{-4}{2\left(1\right)}$

simplify, 2

Plugin $x_v = 2$ to find the $y_v$ value

$y_v=2^{2}-4(2)-5$

$y_v=-9$

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).