# How do you find the vertex and intercepts for x^2 = -8y?

Jul 17, 2017

Only one intercept and that is at the point $\to \left(x , y\right) = \left(0 , 0\right)$
The vertex is at $\left(x , y\right) = \left(0 , 0\right)$

A lot of step by step explanation given.

#### Explanation:

Multiply both sides by (-1). Makes $- 8 y$ positive

$8 y = - {x}^{2}$

Divide both sides by 8

$y = - {x}^{2} / 8$
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Compare to the standard form of $y = a {x}^{2} + b x + c$

All of these $a , b ' s \mathmr{and} c ' s$ change the basis of $y = {x}^{2}$ in particular ways. Lets consider them one at a time.

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$\textcolor{b l u e}{\text{Point 1 - general shape is } \cap}$

Compare $- {x}^{2} / 8$ to $a {x}^{2}$

$a \to - \frac{1}{8}$

As this is negative the graph is of general shape $\cap$ so it has a maximum.

$\textcolor{p u r p \le}{\text{As "a" is a fraction it widens the graph of } y = - {x}^{2}}$
$\textcolor{p u r p \le}{\text{If "a>1" then it make the graph of "y=-x^2" narrower.}}$
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$\textcolor{b l u e}{\text{Point 2 - the y-axis is the axis of symmetry}}$

The part of the equation $+ b x$ moves the graph left or write.

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

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

But our equation is $y = - \frac{1}{8} {x}^{2} \text{ "->" } y = - \frac{1}{8} {x}^{2} + 0 x + 0$

So b=0" "->" "x_("vertex")=(-1/2)xx0/(" "-1/8" ") = 0

As ${x}_{\text{vertex}} = 0$ the graph is symmetrical about the y-axis

So at this point we can sate that Vertex$\to \left(x , y\right) = \left(0 , {y}_{\text{vertex}}\right)$
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$\textcolor{b l u e}{\text{Determine the y-intercept}}$

The constant $c$ in $y = a {x}^{2} + b c + c$ is the y-intercept.

And we have $y = - \frac{1}{8} x + 0 x + 0$ so $c = 0$

As the graph has the y-axis is the axis of symmetry the y-intercept can only be at the vertex.

Vertex$\to \left(x , y\right) = \left(0 , 0\right)$

Graph of $y = - {x}^{2}$ in red. Notice the way that $y = - \frac{1}{8} {x}^{2}$ in blue is wider