Question

Will award a lot of points :)

Answer 1

Answer:

Step-by-step explanation:

Your final answer is the standard form of a parabola.  Since your equation has an x-squared term in it and not a y-squared term, your form will be

$(x-h)^2=4p(y-k)$

To get it into this form we will solve the quadratic for y and then set it equal to 0 so we can complete the square on it.  Solving for y then setting y equal to 0:

$x^2-2x-23=8y$ so

$\frac{1}{8}x^2-\frac{2}{8}x-\frac{23}{8}=0$

We only need to complete the square on the x-terms, so we will move the constant back over to the other side of the equals sign:

$\frac{1}{8}x^2-\frac{2}{8}x=\frac{23}{8}$

The rule for completing the square is that the leading coefficient HAS to be a 1.  Ours is 1/8, so we have to factor it out.  When we do that we are left with:

$\frac{1}{8}(x^2-2x)=\frac{23}{8}$

To complete the square on the left, we take half the linear term, square it, and add it onto both sides.  Our linear term is 2 (the number stuck to the x-term).  Half of 2 is 1, and 1 squared is 1.  So we add it into the parenthesis on the left.  BUT we cannot discount the 1/8 sitting out front there.  It is a multiplier.  So what we actually added on the left is 1/8(1).  That looks like this:

$\frac{1}{8}(x^2-2x+1)=\frac{23}{8}+\frac{1}{8}$

Now we will write the left side into its perfect square binomial (which was the whole reason for doing this!) and simplify the right at the same time:

$\frac{1}{8}(x-1)^2=\frac{24}{8}$

Now we will set the whole thing back to equal y:

$\frac{1}{8}(x-1)^2-3=y$

That's one form.  But you need it in vertex form, so we add 3 to both sides:

$\frac{1}{8}(x-1)^2=y+3$ and then multiply both sides by 8:

$(x-1)^2=8(y+3)$

If you need to break it down further to include what your p value is, then:

$(x-1)^2=4(2)(y+3)$

Either that one or the one right above it should work.