Question

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

Answer 1

`y=2(x +11/4)^2-65/2`

Explanation:

Vertex form:

`y=a(x+h)^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=2x^2 +11x-6`

`y+6=2x^2 +11x`

in the form `ax^2+bx+c` to complete the square a must be 1 and:

`c=(b/2)^2`

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

`y+6=2x^2 +11x`

`y+6=2(x^2 +11/2x)`

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

`y+6 +2c=2(x^2 +11/2x+c)`

now solve fo c:

`c=(b/2)^2=((11/2)/2)^2=121/16`

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

`y+6 +2(121/16)=2(x^2 +11/2x+121/16)`

now complete the square:

`y+6 +53/2=2(x +(11/2)/2)^2`

`y+12/2 +53/2=2(x +11/4)^2`

`y+65/2=2(x +11/4)^2`

`y=2(x +11/4)^2-65/2`

vertex `=(-11/4,-65/2)`

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

Answer 2

vertex `(-11/4, - 169/8)`

Explanation:

`y(x) = 2x^2 + 11x - 6`
The x-coordinate of the vertex is given by the formula
`x = - b/(2a) = - 11/4`
The y-coordinate of vertex is the value of `y(-11/4)` -->
`y(-11/4) = 121/8 - 121/4 - 6 = - 169/8`
Vertex `(-11/4, -169/8)`