What is the vertex of  y=-7x^2-2x+3 ?

Nov 21, 2015

$\left(- \frac{1}{7} , \frac{22}{7}\right)$

Explanation:

We must complete the square to put the equation into vertex form: $y = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is the vertex.

y=-7(x^2+2/7x+color(red)(?))+3

We must complete the square. In order to do this, we must recall that ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, so the middle term, $\frac{2}{7} x$, is $2 x$ times some other number, which we can determine to be $\frac{1}{7}$. Thus, the final term must be ${\left(\frac{1}{7}\right)}^{2}$.

$y = - 7 \left({x}^{2} + \frac{2}{7} x + \textcolor{red}{\frac{1}{49}}\right) + 3 + \textcolor{red}{\frac{1}{7}}$

Note that we had to balance the equation—we can add numbers randomly. When the $\frac{1}{49}$ was added, we must realize that it is actually being multiplied by $- 7$ on the outside of the parentheses, so it is actually like adding $- \frac{1}{7}$ to the right side of the equation. In order to balance the equation we add a positive $\frac{1}{7}$ to the same side.

Now, we can simplify:

$y = - 7 {\left(x + \frac{1}{7}\right)}^{2} + \frac{22}{7}$

Since the vertex is $\left(h , k\right)$, we can determine its location is $\left(- \frac{1}{7} , \frac{22}{7}\right)$. (Don't forget the $h$ value switches signs.)