# What is the vertex form of y=5x^2+4x+7?

Feb 27, 2016

$y = 5 {\left(x + \frac{2}{5}\right)}^{2} + \frac{31}{5}$, where vertex is $\left(- \frac{2}{5} , \frac{31}{5}\right)$

#### 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 $y = 5 {x}^{2} + 4 x + 7$, one should first take $5$ out out of first two terms and then make it complete square, as follows:

$y = 5 {x}^{2} + 4 x + 7 = 5 \left({x}^{2} + \frac{4}{5} x\right) + 7$

To make $\left({x}^{2} + \frac{4}{5} x\right)$, complete square, one has to add and subtract, 'square of half the coefficient of $x$, and thus this becomes

$y = 5 {x}^{2} + 4 x + 7 = 5 \left({x}^{2} + \frac{4}{5} x + {\left(\frac{2}{5}\right)}^{2}\right) + 7 - 5 \cdot {\left(\frac{2}{5}\right)}^{2}$ or

$y = 5 {\left(x + \frac{2}{5}\right)}^{2} + 7 - \frac{4}{5}$ or

$y = 5 {\left(x - \left(- \frac{2}{5}\right)\right)}^{2} + \frac{31}{5}$, where vertex is $\left(- \frac{2}{5} , \frac{31}{5}\right)$