# How do you write a quadratic equation with vertex (2,1) y intercept (0,-3)?

Apr 20, 2017

$y = - {x}^{2} + 4 x - 3$

#### 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) are the coordinates of the vertex and a is a constant.

$\text{here } \left(h , k\right) = \left(2 , 1\right)$

$\Rightarrow y = a {\left(x - 2\right)}^{2} + 1$

$\text{to find a, substitute " (0,-3)" into the equation}$

$- 3 = 4 a + 1$

$\Rightarrow 4 a = - 4 \Rightarrow a = - 1$

$\Rightarrow y = - {\left(x - 2\right)}^{2} + 1 \leftarrow \textcolor{red}{\text{ in vertex form}}$

$\text{distributing and simplifying gives}$

$y = - \left({x}^{2} - 4 x + 4\right) + 1$

$\textcolor{w h i t e}{y} = - {x}^{2} + 4 x - 4 + 1$

$\Rightarrow y = - {x}^{2} + 4 x - 3 \leftarrow \textcolor{red}{\text{ in standard form}}$
graph{-x^2+4x-3 [-10, 10, -5, 5]}