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igomit [66]
3 years ago
15

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

Mathematics
1 answer:
Elena L [17]3 years ago
3 0
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}
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