# How do you find the axis of symmetry, and the maximum or minimum value of the function f(x)=3(x+2)(x-2)?

May 31, 2016

$\textcolor{g r e e n}{\text{Vertex} \to \left(x , y\right) \to \left(0 , 12\right)}$

$\textcolor{g r e e n}{\text{Axis of symmetry } \to x = 0}$

$\textcolor{g r e e n}{\text{The vertex is a minimum}}$

#### Explanation:

Example: suppose we had ${a}^{2} - {b}^{2}$ this is the same as $\left(a - b\right) \left(a + b\right)$

Note that $\left(x - 2\right) \left(x + 2\right)$ is of the same structure

Multiply out the brackets

$\text{ "3(x^2-2^2) = 3x^2+12" } \leftarrow 3 {\left(- 2\right)}^{2} = 3 \times 4 = 12$

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$\textcolor{b l u e}{\text{Observation 1}}$

The $3 {x}^{2}$ is positive so the general shape of the graph is $\cup$ thus there is a $\textcolor{g r e e n}{\text{minimum}}$.

If it had been $- 3 {x}^{2}$ then the general shape would have been $\cap$

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$\textcolor{b l u e}{\text{Observation 2}}$

This is a quadratic equation so the $\textcolor{g r e e n}{\text{axis of symmetry is the "x" coordinate of the vertex}}$
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$\textcolor{b l u e}{\text{Determine the y intercept}}$

Consider the standard form of $y = a {x}^{2} + b x + c$

In the question's equation ther is no $b x$ term but there is a $c$ term in that $c = + 12$

$\textcolor{g r e e n}{{y}_{\text{intercept}} = c = + 12}$
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$\textcolor{b l u e}{\text{Observation 3") ->("Determine x vertex}}$

The $b x$ term in $y = a {x}^{2} + b x + c$ moves the graph left or right from the y axis. As there is no $b x$ term the graph is symmetrical about the y-axis.

$\textcolor{g r e e n}{\text{Thus "x_("vertex")-> x=0)" "color(red)(larr" axis of symmetry}}$

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$\textcolor{b l u e}{\text{Observation 4") ->("Determine y vertex}}$

From above the graph is symmetrical about the y-axis. Consequently y-vertex must be at the point where the graph crosses the y axis, which is:

color(green)(y_("intercept")=c=+12)" "color(red)(larr" " y_("vertex")

To prove my point; substitute $x = 0$ into $y = 3 {x}^{2} + 12$

$\textcolor{g r e e n}{{y}_{\text{vertex}} = 3 \left({0}^{2}\right) + 12 = 12}$

$\textcolor{g r e e n}{\text{Vertex} \to \left(x , y\right) \to \left(0 , 12\right)}$