Question

What are the X intercepts of a parabola with vertex (6,27) and y intercept of (0,-81)?

Answer 1

so hmm let's see if we can get the parabola's equation

if we assume a vertical parabola, "y" in x-terms.... then


`\bf \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\`


`\bf \textit{we know that } \begin{cases} h=6\\ k=27 \end{cases}\implies y=a(x-6)^2+27 \\\\\\ \textit{we also know the y-intercept } \begin{cases} x=-81\\ y=0 \end{cases}\implies 0=a(-81-6)^2+27 \\\\\\ -27=a(-87)^2\implies \cfrac{-27}{(-87)^2}=a\implies -\cfrac{3}{841}=a\\\\ -------------------------------\\\\ \boxed{y=-\cfrac{3}{841}(x-6)^2+27}`

now, to get the x-intercepts, we simply set y = 0


`\bf 0=-\cfrac{3}{841}(x-6)^2+27\implies -27=-\cfrac{3}{841}(x-6)^2 \\\\\\ \cfrac{-27\cdot 841}{-3}=(x-6)^2\implies 7569=(x-6)^2\implies \pm√(7569)=x-6 \\\\\\ \pm 87=x-6\implies \pm 87+6=x\implies x= \begin{cases} 93\\ -81 \end{cases}`

and... since the y = 0, then (93, 0) and (-81, 0)