# What is the vertex form of  y= (6x-6)(x+2)+4x^2+5x?

Nov 23, 2017

Vertex form of equation is $y = 10 {\left(x + 0.55\right)}^{2} - 15.025$

#### Explanation:

$y = \left(6 x - 6\right) \left(x + 2\right) + 4 {x}^{2} + 5 x$ or

$y = 6 {x}^{2} + 12 x - 6 x - 12 + 4 {x}^{2} + 5 x$ or

$y = 10 {x}^{2} + 11 x - 12 \mathmr{and} y = 10 \left({x}^{2} + \frac{11}{10} x\right) - 12$ or

$y = 10 \left\{{x}^{2} + \frac{11}{10} x + {\left(\frac{11}{20}\right)}^{2}\right\} - 10 \cdot {\left(\frac{11}{20}\right)}^{2} - 12$ or

$y = 10 {\left(x + \frac{11}{20}\right)}^{2} - 3.025 - 12$ or

$y = 10 {\left(x + 0.55\right)}^{2} - 15.025$ .Comparing with standard vertex form

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

here $h = - 0.55 , k = - 15.025$ So vertex is at

$\left(- 0.55 , - 15.025\right)$ and vertex form of equation is

$y = 10 {\left(x + 0.55\right)}^{2} - 15.025$ [Ans]