# 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?

Dec 7, 2016

This equation is a parabola. It opens down, and has a vertex of $\left(- 4 , - 7\right) .$

#### 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 $y + {x}^{2} = - \left(8 x + 23\right)$, we see that only $x$ appears squared. Based on the above reasoning, this must be a parabola. Solving for $y$, we get

$y = - {x}^{2} - 8 x - 23$

Factor the coefficient for ${x}^{2}$ out of both $x$ terms:

$\textcolor{w h i t e}{y} = - \left({x}^{2} + \textcolor{red}{8} x\right) - 23$
Complete the square by adding ${\left(\frac{\textcolor{red}{8}}{2}\right)}^{2} = {4}^{2} = \textcolor{m a \ge n t a}{16}$ inside the $x$-brackets. Don't forget to subtract it too.

$\textcolor{w h i t e}{y} = - \left(\underline{{x}^{2} + 8 x + \textcolor{m a \ge n t a}{16}} - \textcolor{m a \ge n t a}{16}\right) - 23$
$\textcolor{w h i t e}{y} = - \left(\underline{{\left(x + 4\right)}^{2}} - 16\right) - 23$

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

$\textcolor{w h i t e}{y} = - {\left(x + 4\right)}^{2} + 16 - 23$
$y = - {\left(x + 4\right)}^{2} - 7$

A parabola in the form $y = a {\left(x - h\right)}^{2} + k$ has a vertex of $\left(h , k\right)$. Here, our parabola has a vertex of $\left(- 4 , - 7\right) .$ 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 "$1 \times a$" units up.
• From this point, go "one unit right" and "$3 \times a$" 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!