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

Aug 15, 2017

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

#### Explanation:

$f \left(x\right) = 2 {x}^{2} + 13 x - 1$
x- coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{13}{4}$
y-coordinate of vertex:
$f \left(- \frac{13}{4}\right) = 2 \frac{169}{16} - 13 \left(\frac{13}{4}\right) - 1 = \frac{336}{16} - \frac{169}{4} - 1 = \frac{352}{16} = - 22$
Vertex form:
f$\left(x\right) = 2 {\left(x + \frac{13}{4}\right)}^{2} - 22$

Aug 15, 2017

$2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{177}{8}$

#### Explanation:

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