Question

Write an equation in standard form of the parabola that has the same shape as the graph of f(x) = -4 x^2, but with vertex at (-19, -18).

Answer 1

`\bf \boxed{y=a(x-{{ h}})^2+{{ k}}}\\ x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -----------------------------\\\\ f(x)=-4x^2\implies \begin{array}{llll} y=-4(x-&0)^2+&0\\ &\uparrow &\uparrow \\ &h&k \end{array} \\\\\\ \textit{now, you're asked for one with a vertex of -19, -18}\\ \textit{that simply means h = -19, and k= -18}`
so, just change those h,k coordinates of the center :)

also, we're asked to put it in "standard form", that simply means, expand the squared binomial, and simplify, you'll end up with a trinomial