Question

How do you identity if the equation `y+x^2=-(8x+23)` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

This equation is a parabola. It opens down, and has a vertex of `(-4,-7).`

Explanation:

• If you have both an `x^2` and a `y^2` term, then you have either an ellipse or a hyperbola.

Move everything to one side. (The other side should be `=0.`)

• If both quadratic terms (`x^2` and `y^2`) have the same sign, then you have an ellipse.

  • If both terms have the same coefficient, then you have a special kind of ellipse called a circle.

• If the quadratic terms have different signs, then you have a hyperbola.

• If only one of your variables (`x` or `y`) is squared, then you have a parabola.

• If the `x` is squared, you have a (classic) parabola that opens up/down.
• If the `y` is squared, you have a parabola that opens right/left.

• If neither `x` or `y` have a squared term, you have a line.

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In the equation `y+x^2=-(8x+23)`, we see that only `x` appears squared. Based on the above reasoning, this must be a parabola. Solving for `y`, we get

`y=-x^2-8x-23`

Factor the coefficient for `x^2` out of both `x` terms:

`color(white)y=-(x^2+color(red)8x)-23`
Complete the square by adding `(color(red)8/2)^2=4^2=color(magenta)16` inside the `x`-brackets. Don't forget to subtract it too.

`color(white)y=-(ul(x^2+8x+color(magenta)(16))-color(magenta)16)-23`
`color(white)y=-(ul((x+4)^2)-16)-23`

Move the 16 outside of the big brackets by multiplying it by the `-1`:

`color(white)y=-(x+4)^2+16-23`
`y=-(x+4)^2-7`

A parabola in the form `y=a(x-h)^2+k` has a vertex of `(h,k)`. Here, our parabola has a vertex of `(-4,-7).` Since `a=-1`, we can graph this parabola by plotting the vertex and then using the 1-3-5 method:

• From the vertex, our next point is at "one unit right" and "`1timesa`" units up.
• From this point, go "one unit right" and "`3timesa`" units up to find the next point.
• Then, we go "one more unit right" and "`5 times a`" units up to get another point.
• Repeat the above steps, going left instead of right.

Note: here, `a=-1`, so our "steps up" will really be "steps down".

And that's it!