Expert: Sherry Wallin - 11/18/2008

Question
Hello homeschool parent and I need ur help step by step how 2 answer these problems for my child I havnt got the answer key yet please answer them god bless



Graph and give me the points for the following parabolas and hyperbolas. For parabolas, list the coordinates  of the vertex and list two other points on the parabola. For hyperbolas, include the asymptotes on the graph, list the coordinates of the vertices, and one other point on the hyperbola

1. x = y^2 - 2

          x^2
2. y = -------
           4

      x^2     y^2
3.  -------- - -------- = 1
       16         9  


4. x^2 + y + 3 = 0

              x^2
5. y^2 - ---------- = 1
               9  

Answer
First off 2 and 4 are parabolas. Do you know what a parabola looks like?
They look sort of like an umbrella that either opens up or down. If you imagine the center of the umbrella, it would be your vertex. There are several ways to determine the vertex but the easiest way is to use -b/2a for the x coordinate and then for the y coordinate you substitute your new found value for x into the equation and solve for y. The standard parabolic equation looks like ax^2+bx+c where a,b,c are coefficients (numbers)and c is also a constant and a <> 0, because if a=0 you would not have a 2nd degree equation. So in problem 2 your a = 1/4 and you have no b or c. The absence of b means it is 0 because you can think of this equation as (1/4)x^2 + 0*x + 0. So the vertex is -0/[2(1/4)] which is of course 0, now plug 0 into (1/4)*0^2 and find that y is also 0 so your vertex is (0,0). You can find any two points on that parabola by choosing some value for x and then substituting it into the equation and solve for y. An easy one is x=1, the (1/4)*1^2 = 1/4 (incidentally 1^2 and (-1)^2 will always give the same value so x =-1 will also give you 1/4. Your two ordered pairs (points on the parabola) are (1,1/4) and (-1,1/4). For #4 solve for y first and then see what what b and c are and proceed as I have shown. #1 is also a parabola but instead it is laying on it's side with it's vertex y = something. Solve for x and use ay^2+by+c as your starting equation. Then it's y coordinate will be -b/2a and you will get the x value by substituting the new found y value into the equation and solving for x.
Both 3 and 5 are hyperbolas.I will work on 3 for you. The first thing you want to do is put the equation into ()^2 - ()^2 = ()^2. Since 16 = 4^2 and 9 = 3^2 you can make this equation as so:(x/4)^2-(y/3)^2 = 1^2. This is an equation of a hyperbola with center at (0,0) and vertices at (-a,0) and (a,0) where a is the square root of the denominator of the first term and in this case x^2 is the first term so the vertices are (-4,0) and (4,0). The denominator of the 2nd term y^2 is called b^2 so b = +-3. To get the asymptotes: the equations for the lines that are asymptotes is
y = (b/a)x and y = -(b/a)x since y is the second term. If x is the second term use y =(a/b)x and y = -(a/b)x. In our problem then the equations for the asymptotes is y = (4/3)x and y = - y = (4/3)x. I have no way to show you the graph unless you want to write another question and send an email address along. In the problem 5 since y^2 is the first term, a is still the denominator of the first term but your vertices are (0,a) and (0,-a).

Math Prof