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

Oct 11, 2017

Vertex form of equation is $y = 4 {\left(x + 0.5\right)}^{2} + 0$

#### Explanation:

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

y = 4(x^2 +x + 0.5^2) -1+1; [4*0.5^2=1] or

$y = 4 {\left(x + 0.5\right)}^{2} + 0$ . Comparing with vertex form of equation

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

$h = - 0.5 \mathmr{and} k = 0$ . So vertex is at $\left(- 0.5 , 0\right)$ and

vertex form of equation is $y = 4 {\left(x + 0.5\right)}^{2} + 0$ [Ans]