# How do you find the vertex for y=2x^2 +11x-6?

Jun 9, 2018

$y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{65}{2}$

#### Explanation:

Vertex form:

$y = a {\left(x + h\right)}^{2} + k$

Vertex #=(-h,k)

To put this in vertex form you must complete the square on the x terms which is a pain because 11 is not divisible by 2:

first isolate the terms with x:

$y = 2 {x}^{2} + 11 x - 6$

$y + 6 = 2 {x}^{2} + 11 x$

in the form $a {x}^{2} + b x + c$ to complete the square a must be 1 and:

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

$a = 2$ so we have to factor it out:

$y + 6 = 2 {x}^{2} + 11 x$

$y + 6 = 2 \left({x}^{2} + \frac{11}{2} x\right)$

now add $c$ to both sides of the equation, we have to add $2 c$ to the left side to account for the 2 we factored out:

$y + 6 + 2 c = 2 \left({x}^{2} + \frac{11}{2} x + c\right)$

now solve fo c:

$c = {\left(\frac{b}{2}\right)}^{2} = {\left(\frac{\frac{11}{2}}{2}\right)}^{2} = \frac{121}{16}$

see what I mean about it being a pain, insert solution in function:

$y + 6 + 2 \left(\frac{121}{16}\right) = 2 \left({x}^{2} + \frac{11}{2} x + \frac{121}{16}\right)$

now complete the square:

$y + 6 + \frac{53}{2} = 2 {\left(x + \frac{\frac{11}{2}}{2}\right)}^{2}$

$y + \frac{12}{2} + \frac{53}{2} = 2 {\left(x + \frac{11}{4}\right)}^{2}$

$y + \frac{65}{2} = 2 {\left(x + \frac{11}{4}\right)}^{2}$

$y = 2 {\left(x + \frac{11}{4}\right)}^{2} - \frac{65}{2}$

vertex $= \left(- \frac{11}{4} , - \frac{65}{2}\right)$

graph{2x^2 +11x-6 [-20.33, 25.28, -25.08, -2.28]}

Jun 9, 2018

vertex $\left(- \frac{11}{4} , - \frac{169}{8}\right)$

#### Explanation:

$y \left(x\right) = 2 {x}^{2} + 11 x - 6$
The x-coordinate of the vertex is given by the formula
$x = - \frac{b}{2 a} = - \frac{11}{4}$
The y-coordinate of vertex is the value of $y \left(- \frac{11}{4}\right)$ -->
$y \left(- \frac{11}{4}\right) = \frac{121}{8} - \frac{121}{4} - 6 = - \frac{169}{8}$
Vertex $\left(- \frac{11}{4} , - \frac{169}{8}\right)$