Question

Create the equation of a quadratic function with a vertex of (5,6) and a y-intercept of -69

Answer 1

if the y-intercept is at -69, meaning the point is (0, -69), thus x = 0, y = -69


`\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ vertex~(5,6)\quad \begin{cases} x=5\\ y=6 \end{cases}\implies y=a(x-5)^2+6 \\\\\\ \textit{we also know that } \begin{cases} x=0\\ y=-69 \end{cases}\implies -69=a(0-5)^2+6 \\\\\\ -75=25a\implies \cfrac{-75}{25}=a\implies a=-3 \\\\\\ therefore\qquad \boxed{y=-3(x-5)^2+6}`