# How do you find the vertex and the intercepts for Y=x^2+5x-6?

Mar 9, 2017

intercepts: $x = - 6 , 1$

vertex: $\left(- \frac{5}{2} , - \frac{49}{4}\right)$.

#### Explanation:

$y = {x}^{2} + 5 x - 6$

For the x-axis intercepts, factor and set to zero:

$y = \left(x + 6\right) \left(x - 1\right) = 0$

$\implies x = - 6 , 1$

For the vertex, complete the square:

$y = {\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} - 6$

$= {\left(x + \frac{5}{2}\right)}^{2} - \frac{49}{4}$

So this quadratic has its minimum value at $\left(- \frac{5}{2} , - \frac{49}{4}\right)$.