# What is the vertex form of y=6x^2+14x-2?

May 30, 2018

$y = 6 {\left(x + \frac{7}{6}\right)}^{2} - \frac{61}{6}$

So your vertex = $\left(- \frac{7}{6} , - \frac{61}{6}\right)$

#### Explanation:

Vertex form is:

$y = a {\left(x + h\right)}^{2} + k$ and the vertex is: #(-h,k)

To put the function in vertex for we have to complete the square with the x values:

$y = 6 {x}^{2} + 14 x - 2$

first isolate the term with x:

$y + 2 = 6 {x}^{2} + 14 x$

to complete the square the following must be done:

$a {x}^{2} + b x + c$

$a = 1$

$c = {\left(\frac{b}{2}\right)}^{2}$

the Square is: ${\left(x + \frac{b}{2}\right)}^{2}$

In your function $a = 6$ so we need to factor that out:

$y + 2 = 6 \left({x}^{2} + \frac{14}{6} x\right)$

$y + 2 = 6 \left({x}^{2} + \frac{7}{3} x\right)$

now add the c in to both sides of the equation, remember on the left we must add in 6c since the c on the right in inside the factored portion:

$y + 2 + 6 c = 6 \left({x}^{2} + \frac{7}{3} x + c\right)$

now solve for c:

$c = {\left(\frac{b}{2}\right)}^{2} = {\left(\frac{\frac{7}{3}}{2}\right)}^{2} = {\left(\frac{7}{6}\right)}^{2} = \frac{49}{36}$

$y + 2 + 6 \left(\frac{49}{36}\right) = 6 \left({x}^{2} + \frac{7}{3} x + \frac{49}{36}\right)$

$y + 2 + \frac{49}{6} = 6 {\left(x + \frac{7}{6}\right)}^{2}$

$y + \frac{61}{6} = 6 {\left(x + \frac{7}{6}\right)}^{2}$

Finally we have vertex form:

$y = 6 {\left(x + \frac{7}{6}\right)}^{2} - \frac{61}{6}$

So your vertex = $\left(- \frac{7}{6} , - \frac{61}{6}\right)$

graph{6x^2+14x-2 [-19.5, 20.5, -15.12, 4.88]}