Question

What is the standard form of the quadratic function that has a vertex at(-3,-5) and goes through the point (0, 13)?O A. y = 2x2 + 12x+13B. y = 2(x + 3)2 - 5O C. y = 2x2 + 12x+18O D. y= x2 + 6x + 9

Answer 1

The vertex of the quadratic is (-3,-5) and the y intercept is (0,13).

The general form of the quadratic is,

$y=ax^2+bx+c$

Substituting the value of (0,13),

$\begin{gathered} 13=a(0^2)+b(0)+c \\ c=13 \end{gathered}$

The vertex is at point (-3,-5),

$\begin{gathered} \frac{-b}{2a}=-3 \\ b=6a\ldots\ldots\text{.}(2) \end{gathered}$

Also (-3,-5) satisfies the equation,

$\begin{gathered} -5=a(-3)^2+b(-3)+c \\ -5=9a-3b+13 \\ 9a-3b=-18 \\ 3a-b=-6 \\ 3a-6a=-6\text{ (from equation (2))} \\ -3a=-6 \\ a=2 \end{gathered}$

from equation (2),

$\begin{gathered} b=6a \\ b=6(2) \\ b=12 \end{gathered}$

Thus, the quadratic can be formed by substituting the values of a, b and c,

$y=2x^2+12x+13$

Thus, option (A) is correct.