Question

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

Answer 1

Only one intercept and that is at the point `->(x,y)=(0,0)`
The vertex is at `(x,y)=(0,0)`

A lot of step by step explanation given.

Explanation:

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

`8y=-x^2`

Divide both sides by 8

`y=-x^2/8`
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Compare to the standard form of `y=ax^2+bx+c`

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

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`color(blue)("Point 1 - general shape is " nn)`

Compare `-x^2/8` to `ax^2`

`a->-1/8`

As this is negative the graph is of general shape `nn` so it has a maximum.

`color(purple)("As "a" is a fraction it widens the graph of "y=-x^2)`
`color(purple)("If "a>1" then it make the graph of "y=-x^2" narrower.")`
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`color(blue)("Point 2 - the y-axis is the axis of symmetry")`

The part of the equation `+bx` moves the graph left or write.

Given that `y=ax^2+bx+c` write as `y=a(x^2+b/ax) +c`

`x_("vertex")=(-1/2)xxb/a`

But our equation is `y=-1/8x^2" "->" "y=-1/8x^2+0x+0`

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

As `x_("vertex")=0` the graph is symmetrical about the y-axis

So at this point we can sate that Vertex`->(x,y)=(0,y_("vertex"))`
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`color(blue)("Determine the y-intercept")`

The constant `c` in `y=ax^2+bc+c` is the y-intercept.

And we have `y=-1/8x+0x+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`->(x,y)=(0,0)`

Graph of `y=-x^2` in red. Notice the way that `y=-1/8x^2` in blue is wider