# What is the the vertex of y=4x^2 + 9x + 15?

Feb 27, 2016

$y = 4 {\left(x - \left(- \frac{9}{8}\right)\right)}^{2} + \frac{159}{16}$, where vertex is $\left(- \frac{9}{8} , \frac{159}{16}\right)$

#### Explanation:

Vertex form of equation is of type y = a(x – h)^2 + k, where $\left(h , k\right)$ is the vertex. For this, in the equation $y = 4 {x}^{2} + 9 x + 15$, one should first take $4$ out out of first two terms and then make it complete square, as follows:

$y = 4 {x}^{2} + 9 x + 15 = 4 \left({x}^{2} + \frac{9}{4} x\right) + 15$

To make $\left({x}^{2} + \frac{9}{4} x\right)$, complete square, one has to add and subtract, 'square of half the coefficient of $x$, and thus this becomes

$y = 4 {x}^{2} + 9 x + 15 = 4 \left({x}^{2} + \frac{9}{4} x + {\left(\frac{9}{8}\right)}^{2}\right) + 15 - 4 \cdot {\left(\frac{9}{8}\right)}^{2}$ or

$y = 4 {\left(x + \frac{9}{8}\right)}^{2} + 15 - \frac{81}{16}$ or

$y = 4 {\left(x - \left(- \frac{9}{8}\right)\right)}^{2} + \frac{159}{16}$, where vertex is $\left(- \frac{9}{8} , \frac{159}{16}\right)$