Question

Find the x intercepts of the parabola with vertex (6,-5) and y- int (0,175)

Answer 1

Hmmm so.. .it could be either a horizontal or vertical parabola, let's assume or use a vertical parabola vertex equation then... namely, the squared variable being "x".

$\bf \qquad \textit{parabola vertex form}\\\\ \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ \begin{cases} h=6\\ k=-5 \end{cases}\implies y=a(x-6)^2-5 \\\\\\ \textit{we also know } \begin{cases} x=0\\ y=175 \end{cases}\implies 175=a(0-6)^2-5\implies 180=a6^2 \\\\\\ \cfrac{180}{36}=a\implies 5=a\qquad thus\qquad \boxed{y=5(x-6)^2-5}$

$\bf \textit{at the x-intercepts, y = 0, thus }\qquad 0=5(x-6)^2-5 \\\\\\ 5=5(x-6)^2\implies \cfrac{5}{5}=(x-6)^2\implies \pm\sqrt{1}=x-6 \\\\\\ \pm 1+6=x\implies x= \begin{cases} 7\\ 5 \end{cases}$