What is the vertex form of y= x^2-3x-10 ?

May 16, 2017

minimum vertex at $\left(\frac{3}{2} , - \frac{49}{4}\right)$

Explanation:

$y = {x}^{2} - 3 x - 10$

using completing a square,

$y = {\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} - 10$

$y = {\left(x - \frac{3}{2}\right)}^{2} - \frac{49}{4}$

since a coeficient of $\left(x - \frac{3}{2}\right)$ has a +ve value, we can say that it has a minimum vertex at $\left(\frac{3}{2} , - \frac{49}{4}\right)$