Question

How do you write `0=2X^2+13X-1` in vertex form?

Answer 1

f(x) = 2(x + 13/4)^2 - 22

Explanation:

`f(x) = 2x^2 + 13x - 1`
x- coordinate of vertex:
`x = - b/(2a) = - 13/4`
y-coordinate of vertex:
`f(- 13/4) = 2(169)/16 - 13(13/4) - 1 = 336/16 - 169/4 - 1 = 352/16 = - 22`
Vertex form:
f`(x) = 2(x + 13/4)^2 - 22`

Answer 2

`2(x+13/4)^2-177/8`

Explanation:

`2x^2+13x-1=0`
To convert standard from of the quadratic equation into the vertex form complete the square:
`2(x^2+(13/2)x)-1=0`
`2[x^2+(13/2)x+(13/4)^2]-1-2(13/4)^2=0`
`2(x+13/4)^2-(1+169/8)=0`
`2(x+13/4)^2-177/8=0` => in the vertex form of `y=a(x-h)^2+k` where: `(h,k)` is the vertex.
Thus in this case:
#(-13/4, -177/8) is the vertex