How do you write the quadratic equation given Vertex: (3,-11) Passing though: (5,2)?

Feb 19, 2017

$y = \frac{13}{4} {x}^{2} - \frac{39}{2} x + \frac{73}{4}$

Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h ,k) is the vertex and a, a constant.

$\text{here } \left(h , k\right) = \left(3 , - 11\right)$

$\Rightarrow y = a {\left(x - 3\right)}^{2} - 11 \text{ is the partial equation}$

To find a, substitute the point (5 ,2) into the equation.

$2 = a {\left(5 - 3\right)}^{2} - 11$

$\Rightarrow 2 = 4 a - 11$

$\Rightarrow a = \frac{2 + 11}{4} = \frac{13}{4}$

$\Rightarrow y = \frac{13}{4} {\left(x - 3\right)}^{2} - 11 \leftarrow \textcolor{red}{\text{ in vertex form}}$

distributing and simplifying gives.

$y = \frac{13}{4} {x}^{2} - \frac{39}{2} x + \frac{73}{4} \leftarrow \textcolor{red}{\text{ in standard form}}$