Let f (x)=x^2+2x-35 enter the x-intercepts of the quadratic function in the boxes __and___
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We will begin by grouping the x terms together and the y terms together so we can complete the square and see what we're looking at. 
. Now we need to move that 36 over by adding to isolate the x and y terms. 
. Now we need to complete the square on the x terms and the y terms. Can't do that, though, til the leading coefficients on the squared terms are 1's. Right now they are 9 and 4. Factor them out: 
. Now let's complete the square on the x's. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so add it into the parenthesis. BUT don't forget about the 9 hanging around out front there that refuses to be forgotten. It is a multiplier. So we are really adding in is 9*4 which is 36. Half the linear term on the y's is 3. 3 squared is 9, but again, what we are really adding in is -4*9 which is -36. Putting that altogether looks like this thus far: 
. The right side simplifies of course to just 36. Since we have a minus sign between those x and y terms, this is a hyperbola. The hyperbola has to be set to equal 1. So we divide by 36. At the same time we will form the perfect square binomials we created for this very purpose on the left: 
. Since the 9 is the bigger of the 2 values there, and it is under the y terms, our hyperbola has a horizontal transverse axis. a^2=4 so a=2; b^2=9 so b=3. Our asymptotes have the formula for the slope of 
which for us is a slope of negative and positive 3/2. Using the slope and the fact that we now know the center of the hyperbola to be (2, 3), we can solve for b and rewrite the equations of the asymptotes. 
give us a b of 0 so that equation is y = 3/2x. For the negative slope, we have 
which gives us a b value of 6. That equation then is y = -3/2x + 6. And there you go!
