How do you write f(x) = -4x^2 - 16x + 3 in vertex form?

Jun 18, 2017

$f \left(x\right) = - 4 {\left(x + 2\right)}^{2} + 19$

Explanation:

$\text{for the standard form of a parabola } y = a {x}^{2} + b x + c$

"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)

$y = - 4 {x}^{2} - 16 x + 3 \text{ is in standard form}$

$\text{with } a = - 4 , b = - 16 , c = 3$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 16}{- 8} = - 2$

$\text{substitute into f(x) for y-coordinate}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - 4 {\left(- 2\right)}^{2} - 16 \left(- 2\right) + 3 = 19$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 2 , 19\right)$

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

• y=a(x-h)^2+k

where ( h , k ) are the coordinates of the vertex and a is a constant.

$\text{here " (h,k)=(-2,19)" and } a = - 4$

$\Rightarrow y = - 4 {\left(x + 2\right)}^{2} + 19 \leftarrow \textcolor{red}{\text{ in vertex form}}$