# How do you write f(x)=-6x^2-5x+1 in vertex form?

Apr 24, 2017

$y = - 6 {\left(x + \frac{5}{12}\right)}^{2} + \frac{49}{24}$

#### Explanation:

Vertex form is $f \left(x\right) = a {\left(x - h\right)}^{2} + k$

We have $f \left(x\right) = - 6 {x}^{2} - 5 x + 1$

= $- 6 \left({x}^{2} + \frac{5}{6} x\right) + 1$

= $- 6 \left({x}^{2} + 2 \times \frac{5}{12} \times x + {\left(\frac{5}{12}\right)}^{2}\right) - \left(- 6\right) \times {\left(\frac{5}{12}\right)}^{2} + 1$

= $- 6 {\left(x + \frac{5}{12}\right)}^{2} + \frac{25}{24} + 1$

= $- 6 {\left(x + \frac{5}{12}\right)}^{2} + \frac{49}{24}$

and vertex is $\left(- \frac{5}{12} , - \frac{49}{24}\right)$

graph{-6x^2-5x+1 [-3.8, 3, -1, 2.4]}