Question

Find the equation of the quadratic function with zeros 10 and 14 and vertex at (12, -8).

Answer 1

we'll do the same we did on the previous one, using the vertex form.

our vertex is at (12, -8) thus h = 12, k = -8

and our zeros are 10, 0 and 14,0, so.. .we'll use say hmm (10,0) to get the "a" coefficient.


`\bf \qquad \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\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ vertex\ (12,-8)\ \begin{cases} h=12\\ k=-8 \end{cases}\implies y=a(x-12)^2-8 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=10 \end{cases}\implies 0=a(10-12)^2-8 \\\\\\ 8=a(-2)^2\implies 8=4a\implies \boxed{2=a} \\\\\\ thus\qquad \boxed{y=2(x-12)^2-8}`