# What is the the vertex of y = (x - 3)^2 + 7x-12 ?

Feb 29, 2016

color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)

#### Explanation:

You have to combine these variables before you can do any thing else.

Given:$\text{ } y = {\left(x - 3\right)}^{2} + 7 x - 12$

$y = {x}^{2} - 6 x + 9 + 7 x - 12$

$y = {x}^{2} + x - 3$...........................(1)

From this point you can 'complete the square' or do a part version of the process.

I am opting for the part process

Write as:

Consider the coefficient of $+ x$ in equation ( 1 ), which is 1
Apply $\left(- \frac{1}{2}\right) \times 1 = - \frac{1}{2}$

${x}_{\text{vertex}} = - \frac{1}{2}$....... Fast isn't it!
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$\textcolor{b r o w n}{\text{A word of warning:}}$ Given the standard form of
$y = a {x}^{2} + b x + c$ you need to convert this to

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

In your case $a = 1$
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Substitute $x = - \frac{1}{2}$ into equation (1)

${y}_{\text{vertex}} = {\left(- \frac{1}{2}\right)}^{2} + \left(- \frac{1}{2}\right) - 3$

${y}_{\text{vertex}} = \frac{1}{4} - \frac{1}{2} - 3 = - 3 \frac{1}{4}$
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color(blue)("Vertex" -> (x,y) -> (-1/2 ,-3 1/4)