?Write the standard form of the equation of the parabola that has a vertex at (-6,-5)and passes through the point (-8,6). ?
A)? $f ( x ) = \frac { 11 } { 36 } ( x - 6 ) ^ { 2 } - 5$
B)? $f ( x ) = - \frac { 5 } { 36 } ( x + 5 ) ^ { 2 } + 8$
C)? $f ( x ) = \frac { 11 } { 4 } ( x + 6 ) ^ { 2 } - 5$
D)? $f ( x ) = \frac { 3 } { 2 } ( x + 6 ) ^ { 2 } - 8$
E)? $f ( x ) = \frac { 5 } { 8 } ( x + 5 ) ^ { 2 } + 6$

Question

Quizplus · Accepted Answer

The standard form of a parabola's equation is $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. For a vertex at $(-6, -5)$, the equation starts as $f(x) = a(x + 6)^2 - 5$. To find $a$, use the point $(-8, 6)$. Substituting $-8$ for $x$ and $6$ for $f(x)$, we get $6 = a(-8 + 6)^2 - 5$. Solving for $a$ gives $a = \frac{11}{4}$, making the correct equation $f(x) = \frac{11}{4}(x + 6)^2 - 5$.

?Write the Standard Form of the Equation of the Parabola f(x)=1136(x−6)2−5f ( x ) = \frac { 11 } { 36 } ( x - 6 ) ^ { 2 } - 5f(x)=3611​(x−6)2−5

?Write the Standard Form of the Equation of the Parabola $f ( x ) = \frac { 11 } { 36 } ( x - 6 ) ^ { 2 } - 5$