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?

1 Answer
Dec 7, 2016

Answer:

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!