Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

write an equation for a prabola with a vertex at the origin, passing through(5,6), and symmetric with respect to the x-axis

Sagot :

a parabola with symmetry or mirror image across the x-axis, will be a horizontal parabola, namely one that opens either to the right or left.

a horizontal parabola will be a parabola whose independent variable is the "y", so it'll be an x(y) type of equation.

[tex]~~~~~~\textit{horizontal parabola vertex form} \\\\ x=a(y- k)^2+ h\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\supset}\qquad \stackrel{"a"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies x=a(y-0)^2+0\qquad \textit{we also know that} \begin{cases} x=5\\ y=6 \end{cases} \\\\\\ 5=a(6-0)^2 +0\implies 5=36a\implies \cfrac{5}{36}=a~\hfill \boxed{x=\cfrac{5}{36}y^2}[/tex]