# What is the vertex form of 5y = 13x^2+20x+42?

Oct 16, 2016

$y = \frac{13}{5} {\left(x - - \frac{10}{13}\right)}^{2} + \frac{446}{65}$

#### Explanation:

Divide both sides by 5:

$y = \frac{13}{5} {x}^{2} + 4 x + \frac{42}{5}$

The equation is in standard form, $y = a {x}^{2} + b x + c$. In this form the x coordinate, h, of the vertex is:

$h = - \frac{b}{2 a}$

$h = - \frac{4}{2 \left(\frac{13}{5}\right)} = - \frac{20}{26} = - \frac{10}{13}$

The y coordinate, k, of the vertex is the function evaluated at h.

$k = \frac{13}{5} {\left(- \frac{10}{13}\right)}^{2} + 4 \left(- \frac{10}{13}\right) + \frac{42}{5}$

$k = \frac{13}{5} \left(- \frac{10}{13}\right) \left(- \frac{10}{13}\right) - \frac{40}{13} + \frac{42}{5}$

$k = \left(- 2\right) \left(- \frac{10}{13}\right) - \frac{40}{13} + \frac{42}{5}$

$k = \frac{20}{13} - \frac{40}{13} + \frac{42}{5}$

$k = - \frac{20}{13} + \frac{42}{5}$

$k = - \frac{100}{65} + \frac{546}{65}$

$k = \frac{446}{65}$

The vertex form of the equation of a parabola is:

$y = a {\left(x - h\right)}^{2} + k$

Substituting in our known values:

$y = \frac{13}{5} {\left(x - - \frac{10}{13}\right)}^{2} + \frac{446}{65}$