How do you write a quadratic equation with vertex (2,-4) and the point (1,-1)?

Mar 30, 2018

$y = 3 {x}^{2} - 12 x + 8$

Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a is}$
$\text{a multiplier}$

$\text{here } \left(h , k\right) = \left(2 , - 4\right)$

$\Rightarrow y = a {\left(x - 2\right)}^{2} - 4$

$\text{to find a substitute "(1,-1)" into the equation}$

$- 1 = a - 4 \Rightarrow a = - 1 + 4 = 3$

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

$\text{distributing and simplifying gives}$

$y = 3 \left({x}^{2} - 4 x + 4\right) - 4$

$\Rightarrow y = 3 {x}^{2} - 12 x + 8 \leftarrow \textcolor{red}{\text{equation in standard form}}$
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