How do you find the vertex of y = x^2-8x+7?

Jul 21, 2015

Complete the square to reformulate the quadratic in vertex form:

$y = {x}^{2} - 8 x + 7 = {\left(x - 4\right)}^{2} + \left(- 9\right)$

So the vertex is at $\left(4 , - 9\right)$

Explanation:

In the general case we can complete the square as follows:

$y = a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - \frac{{b}^{2}}{4 a}\right)$

Notice especially the $\frac{b}{2 a}$ term, which gives us the $- 4$. Normally you just observe the $a$ and $b$ terms and look at what you get when you multiply out $a {\left(x - \frac{b}{2 a}\right)}^{2}$ rather than following the above formula. On this occasion, let's just use the formula...

In our case $a = 1$, $b = - 8$ and $c = 7$, so

$y = {x}^{2} - 8 x + 7$

$= 1 \cdot {\left(x + \frac{- 8}{2 \cdot 1}\right)}^{2} + \left(7 - \frac{{\left(- 8\right)}^{2}}{4 \cdot 1}\right)$

$= {\left(x - 4\right)}^{2} + \left(7 - \frac{64}{4}\right)$

$= {\left(x - 4\right)}^{2} + \left(7 - 16\right)$

$= {\left(x - 4\right)}^{2} + \left(- 9\right)$

Vertex form of a quadratic is:

$y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex.

So in our case $\left(h , k\right) = \left(4 , - 9\right)$

Why is this the vertex?

Well ${\left(x - h\right)}^{2} \ge 0$ will only be zero when $x - h = 0$ - that is when $x = h$. When $x = h$ then $y = 0 + k = k$, hence $\left(h , k\right)$.