Write the standard form of the equation of the parabola that has a vertex at $\left( \frac { - 2 } { 3 } , \frac { 1 } { 9 } \right)$ and passes through the point $( 3 , - 4 )$.
A)
$$f ( x ) = - \frac { 37 } { 11 } \left( x + \frac { 2 } { 3 } \right) ^ { 2 } + \frac { 1 } { 9 }$$
B)
$$f ( x ) = - \frac { 37 } { 121 } \left( x - \frac { 3 } { 2 } \right) ^ { 2 } + \frac { 1 } { 9 }$$
C)
$$f ( x ) = - \frac { 37 } { 121 } \left( x + \frac { 2 } { 3 } \right) ^ { 2 } + \frac { 1 } { 9 }$$
D)
$$f ( x ) = - \frac { 37 } { 11 } \left( x - \frac { 2 } { 3 } \right) ^ { 2 } - \frac { 1 } { 9 }$$
E)
$$f ( x ) = - \frac { 37 } { 25 } \left( x - \frac { 3 } { 2 } \right) ^ { 2 } - \frac { 1 } { 9 }$$

Question

Quizplus · Accepted Answer

The standard form of the equation of a parabola is given by 
$$y=a(x-h)^2+k$$
where the vertex is at $(h,k)$.
Using the information given, we have 
$$h=-\frac{2}{3}$$
$$k=\frac{1}{9}$$
Substituting these values, we get 
$$y=a(x+\frac{2}{3})^2+\frac{1}{9}$$
To find the value of $a$, we use the fact that the parabola passes through the point $(3,-4)$. Substituting these values, we get
$$-4=a(3+\frac{2}{3})^2+\frac{1}{9}$$
Simplifying,
$$-37=a\cdot\frac{100}{9}$$
$$a=-\frac{333}{100}$$
Substituting back in the equation, we get 
$$y=-\frac{333}{100}(x+\frac{2}{3})^2+\frac{1}{9}$$
Multiplying both sides by $100$ (to get rid of the fraction), we get
$$100y=-\frac{333}{25}(x+\frac{2}{3})^2+1$$
Which is the standard form of the equation of the parabola. Thus, the correct choice is $\boxed{\textbf{(C)}}$.

Write the Standard Form of the Equation of the Parabola (−23,19)\left( \frac { - 2 } { 3 } , \frac { 1 } { 9 } \right)(3−2​,91​)

Write the Standard Form of the Equation of the Parabola $\left( \frac { - 2 } { 3 } , \frac { 1 } { 9 } \right)$