Solving quadratics using $y=a(x-p)^2+q$ ?

Question

The vertex is $(-4, 2)$ and the y-int is $(-3,-1)$. How would I solve this, or find out what "$a$" is so I can write the equation and graph it?

Answer 1

You have the equation $$ y = a(x-p)^2 + q $$ Note that the vertex for this quadratic is at $(p, q)$, so we know that $p = -4$ and $q = 2$. This gives us: $$ y = a(x+4)^2 + 2 $$ Now, plugging in our point $(-3, -1)$, we get: $$ -1 = a(-3+4)^2 + 2 \Rightarrow a = -3 $$ Therefore our equation is: $$ y = -3(x+4)^2 + 2 $$

Answer 2

The vertex form of a parabola tells you that $(p,q)$ is the vertex of the parabola (so $p=-4, q=2$). Now, plug in the other point, and solve for $a$.