# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function x^2 = 12y?

Feb 14, 2016

Axis of symmetry is y-axis
Minimum at $\left(x , y\right) = \left(0 , 0\right)$

#### Explanation:

Given:|$\text{ } {x}^{2} = 12 y$

Divide both sides by 12 ($\div 12 \text{ is the same as } \times \frac{1}{12}$)

$\frac{1}{12} {x}^{2} = \frac{12}{12} \times y$

But $\frac{12}{12} = 1$ giving

$y = \frac{1}{12} {x}^{2}$
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$\textcolor{b l u e}{\text{Let us think about what the numbers are actually doing}}$

$\textcolor{b r o w n}{\text{Point 1}}$

If you had $y = {x}^{2}$ the graph would be of the shape type U and would have the y-axis in the middle of the curve. Thus the axis of symmetry for that case is the y-axis

$\textcolor{b r o w n}{\text{Point 2}}$

So what is the $\frac{1}{12}$ doing? It is a modifies the graph of $y = {x}^{2}$

It is saying: take any value of $x$. Move along the x-axis such that you a looking at $\frac{1}{12}$ of the value of $x$. See where on the curve of $y = {x}^{2}$ the point would be and move that point horizontally to be above the actual value of $x$

So a coefficient (k) of ${x}^{2}$ such that $0 < k < 1$ widens the standard curve of $y = {x}^{2}$ and any (k) such that $1 < k$ narrows the standard curve of $y = {x}^{2}$

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$\textcolor{b r o w n}{\text{Point 3 - Conclusion}}$

$y = \frac{1}{12} {x}^{2}$ is almost the same as $y = {x}^{2}$ but wider.

They both have the same axis of symmetry: y-axis

They both have the same minimum: $\text{min } \to \left(x , y\right) \to \left(0 , 0\right)$
because they both have the basic shape of U
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Note that if we had $y = - \frac{1}{12} {x}^{2}$ then the graph shape would be of type $\bigcap$ and thus we would have a maximum. 