What is the the vertex of y = (x -4)^2+9x-20 ?

Dec 3, 2015

$y = {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{4}$

Vertex $\left(- \frac{1}{2} , - \frac{17}{4}\right)$
Axis of symmetry: $x = - \frac{1}{2}$

Explanation:

Remember quadratic form is $y = a {x}^{2} + b x + c \text{ " " " " " } \left(1\right)$

The vertex form of the quadratic equation is $y = a {\left(x - h\right)}^{2} + k \text{ " " " " " " " " " " } \left(2\right)$

Notice the equation is $y = {\left(x - 4\right)}^{2} + 9 x - 20$ is not in the vertex form.

We begin by expand the equation like so

$y = {\left(x - 4\right)}^{2} + 9 x - 20$

$y = \left(x - 4\right) \left(x - 4\right) + 9 x - 20$
$y = \left({x}^{2} - 4 x - 4 x + 16\right) + 9 x - 20$
$y = {x}^{2} - 8 x + 16 + 9 x - 20$
$y = {x}^{2} + x - 4 \text{ " " " " " " " " " " " " } \left(3\right)$

After we simplify function, we have $y = {x}^{2} + x - 4$ , we can write it in the vertex form by the process of completing the square.

$y = \left({x}^{2} + x + \textcolor{red}{\square}\right) - 4 - \textcolor{b l u e}{\square} \text{ " " } \left(4\right)$

Note: the goal of completing the square if to create a perfect trinomial.
The number in the square is ${\left(\frac{b}{2}\right)}^{2}$

In this case, the middle term is $1$ , ${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4} \text{ " " " } \left(5\right)$

$y = \left({x}^{2} + x + \textcolor{red}{\frac{1}{4}}\right) - 4 - \textcolor{b l u e}{\frac{1}{4}} \text{ " " } \left(6\right)$

$y = \left({x}^{2} + x + \textcolor{red}{\frac{1}{4}}\right) - \frac{16}{4} - \textcolor{b l u e}{\frac{1}{4}}$
$y = {\left(x + \frac{1}{2}\right)}^{2} - \frac{17}{4} \text{ " " " " } \left(7\right)$

There is alternative method to fid the vertex using
${x}_{v e r t e x} = - \frac{b}{2 a}$

${y}_{v e r t e x} = f \left(- \frac{b}{2 a}\right)$

I hope this help.