# What is the vertex form of  7y = - 13x^2 -15x + 2 ?

Feb 13, 2016

$y = - \frac{13}{7} {\left(x + \frac{15}{26}\right)}^{2} + \frac{329}{364}$

#### Explanation:

First, get the equation into its typical form by dividing both sides by $7$.

$y = - \frac{13}{7} {x}^{2} - \frac{15}{7} x + \frac{2}{7}$

Now, we want to get this into vertex form:

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

First, factor the $- \frac{13}{7}$ from the first two terms. Note that factoring a $- \frac{13}{7}$ from a term is the same as multiplying the term by $- \frac{7}{13}$.

$y = - \frac{13}{7} \left({x}^{2} + \frac{15}{13} x\right) + \frac{2}{7}$

Now, we want the term in the parentheses to be a perfect square. Perfect squares come in the pattern ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$.

Here, the middle term $\frac{15}{13} x$ is the middle term of the perfect square trinomial, $2 a x$. If we want to determine what $a$ is, divide $\frac{15}{13} x$ by $2 x$ to see that $a = \frac{15}{26}$.

This means that we want to add the missing term in the parentheses to make the group equal to ${\left(x + \frac{15}{26}\right)}^{2}$.

y=-13/7overbrace((x^2+15/13x+?))^((x+15/26)^2)+2/7

The missing term at the end of the perfect square trinomial is ${a}^{2}$, and we know that $a = \frac{15}{26}$, so ${a}^{2} = \frac{225}{676}$.

Now we add $\frac{225}{676}$ to the terms in the parentheses. However, we can't go adding numbers to equations willy-nilly. We must balance what we just added on the same side of the equation. (For example, if we had added $2$, we would need to add $- 2$ to the same side of the equation for a net change of $0$).

y=color(blue)(-13/7)(x^2+15/13x+color(blue)(225/676))+2/7+color(blue)?

Notice that we haven't actually added $\frac{225}{676}$. Since it's inside of the parentheses, the term on the outside is being multiplied in. Thus, the $\frac{225}{676}$ actually has a value of

$\frac{225}{676} \times - \frac{13}{7} = \frac{225}{52} \times - \frac{1}{7} = - \frac{225}{364}$

Since we have actually added $- \frac{225}{364}$, we must add a positive $\frac{225}{364}$ to the same side.

$y = - \frac{13}{7} {\left(x + \frac{15}{26}\right)}^{2} + \frac{2}{7} + \frac{225}{364}$

Note that $\frac{2}{7} = \frac{104}{364}$, so

color(red)(y=-13/7(x+15/26)^2+329/364

This is in vertex form, where the parabola's vertex is at $\left(h , k\right) \to \left(- \frac{15}{26} , \frac{329}{364}\right)$.

We can check our work by graphing the parabola:

graph{7y = - 13x^2 -15x + 2 [-4.93, 4.934, -2.466, 2.466]}

Note that $- \frac{15}{26} = - 0.577$ and $\frac{329}{364} = 0.904$, which are the values obtained by clicking on the vertex.