Question 223764
You can write a quadratic equation in "vertex form":
{{{y-k=a(x-h)^2}}}, where the vertex is (h, k).


The answer will still have "x" and "y" in the equation, so we just need to find the "a", "h" and "k".
You already have the "h" and "k" (from the vertex).  So, let's plug them in:
{{{y-(8)=a(x-(-1))^2}}}
{{{y-8=a(x+1)^2}}} (simplified)


Looking good, but we still need "a".  Now, where in the world are we going to find THAT???


Wait.  I have an idea.  They said the y-intercept is 7.  In other words, it crosses the y-axis at y=7.  Hey!  They just gave us another ordered pair:  (0, 7).


Let's plug that in:
{{{7-8=a(0+1)^2}}} (plugged in)
{{{-1=a(1)^2}}}
{{{-1=a(1)}}}
{{{-1=a}}}


Now we have all three pieces of our puzzle:
a=-1
h=-1
k=8


Plug in the pieces:
{{{y-8=-1(x+1)^2}}} (Notice that the answer still has "x" and "y" in the equation.)


Oops!  They want the answer in standard form.  No problem.
{{{y-8=-1(x+1)^2}}}
{{{y-8=-1(x+1)*(x+1)}}} (squared the (x+1) )
{{{y-8=-1(x^2+2*x+1)}}} (F.O.I.L.)
{{{y-8=-x^2-2*x-1)}}} (distributed the -1)
{{{y=-x^2-2*x+7)}}} (added 8 to both sides)


Let's graph it just to be sure the vertex is at (-1, 8) and y-intercept is at 7.


{{{graph(600, 400, -10, 10, -10, 10, -x^2+-2*x+7)}}}


Shazam!  It checks out.


<hr />For more worked problems, or tutoring, please visit <a href="http://www.mathlocker.com">MathLocker.com</a>.<hr />