# What is the vertex form of y=-4x^2-4x+1?

Apr 12, 2018

The vertex form of equation is $y = - 4 {\left(x + \frac{1}{2}\right)}^{2} + 2$

#### Explanation:

$y = - 4 {x}^{2} - 4 x + 1$ or

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

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

$y = - 4 {\left(x + \frac{1}{2}\right)}^{2} + 2$ . Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = - \frac{1}{2} , k = 2 \therefore$ Vertex is at $\left(- 0.5 , 2\right)$

The vertex form of equation is $y = - 4 {\left(x + \frac{1}{2}\right)}^{2} + 2$

graph{-4x^2-4x+1 [-10, 10, -5, 5]}

Apr 12, 2018

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

#### Explanation:

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

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

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

$\text{using the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

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

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} + x$

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

$\textcolor{w h i t e}{\Rightarrow y} = - 4 {\left(x + \frac{1}{2}\right)}^{2} - 4 \left(- \frac{1}{4} - \frac{1}{4}\right)$

$\textcolor{w h i t e}{\Rightarrow y} = - 4 {\left(x + \frac{1}{2}\right)}^{2} + 2 \leftarrow \textcolor{red}{\text{in vertex form}}$