# What is the vertex of y=3(x-2)^2+1?

Jan 6, 2016

$\text{vertex} \to \left(x , y\right) \to \left(2 , 1\right)$

#### Explanation:

$\textcolor{b r o w n}{\text{Introduction to idea of method.}}$

When the equation is in the form $a {\left(x - b\right)}^{2} + c$ then ${x}_{\text{vertex}} = \left(- 1\right) \times \left(- b\right)$

If the equation form had been $a {\left(x + b\right)}^{2} + c$ then ${x}_{\text{vertex}} = \left(- 1\right) \times \left(+ b\right)$
color(brown)(underline(color(white)(" ."))

color(blue)("To find "x_("vertex"))

So for $y = 3 {\left(x - 2\right)}^{2} + 1 :$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- 1\right) \times \left(- 2\right) = + 2}$
color(brown)(underline(color(white)(" ."))

color(blue)("To find "y_("vertex"))
Substitute +2 into the original equation to find ${y}_{\text{vertex}}$

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

$\textcolor{b l u e}{{y}_{\text{vertex}} = {0}^{2} + 1 = 1}$

color(brown)("Also notice this value is the same as the constant of +1 that is in the" $\textcolor{b r o w n}{\text{vertex form equation.}}$
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Thus: $\textcolor{g r e e n}{\text{vertex} \to \left(x , y\right) \to \left(2 , 1\right)}$

$\textcolor{p u r p \le}{\text{~~~~~~~~~~~~~~~~~~~ Foot note ~~~~~~~~~~~~~~}}$

Suppose the equation had been presented in the form of:

$y = 3 {x}^{2} - 12 x + 13$

write as $y = 3 \left({x}^{2} - 4 x\right) + 13$

If we carry out the mathematical process of
$\left(- \frac{1}{2}\right) \times \left(- 4\right) = + 2 = {x}_{\text{vertex}}$

The -4 comes from the $- 4 x \text{ in } \left({x}^{2} - 4 x\right)$

$\textcolor{p u r p \le}{\text{ ~~~~~~~~~~~~~~~~~~End Foot note~~~~~~~~~~~~~~~~~~~~~~~~}}$