Question 823880
A key to a quick solution is noticing that the points (-2, -10) and (1, -10) have the same y-coordinate. Because of the symmetry of parabolas, these two points tells us that the axis of symmetry and the vertex are halfway between -2 and 1:
{{{(-2+1)/2 = -1/2}}}
So the x-coordinate of the vertex is -1/2.<br>
We will use this to build the vertex form, {{{y = a(x-h)^2 + k}}} for the equation of this parabola. (Then we will transform the equation into standard {{{y = ax^2+bx+c}}} form.) In the vertex form the "h" and the "k" are the x and y coordinates of the vertex. Since we have already found the x-coordinate we can start with:
{{{y = a(x-(-1/2))^2 + k}}}
which simplifies to:
{{{y = a(x+1/2)^2 + k}}}<br>
Next we will substitute in the coordinates of the given points. I'll use (1, -10):
{{{(-10) = a((1)+1/2)^2 + k}}}
Simplifying...
{{{-10 = a(3/2)^2 + k}}}
{{{-10 = a(9/4) + k}}}
{{{-10 = (9/4)a + k}}}
Solving this for k:
{{{-(9/4)a - 10 = k}}}<br>
Now we'll repeat this with another point. I'll use (-3, -6):
{{{(-6) = a((-3)+1/2)^2 + k}}}
Simplifying...
{{{-6 = a(-5/2)^2 + k}}}
{{{-6 = a(25/4) + k}}}
{{{-6 = (25/4)a + k}}}<br>
Now we'll substitute, into this equation, the expression we got earlier for k:
{{{-6 = (25/4)a + (-(9/4)a - 10)}}}
With only the "a" left, we can solve for it. Simplifying...
{{{-6 = (16/4)a - 10)}}}
{{{-6 = 4a - 10)}}}
Adding 10:
{{{4 = 4a)}}}
Dividing by 4:
1 = a<br>
Now we can use this to find k:
{{{-(9/4)a - 10 = k}}}
{{{-(9/4)(1) - 10 = k}}}
{{{-9/4 - 40/4 = k}}}
{{{-49/4 = k}}}<br>
Our vertex form is now complete:
{{{y = (1)(x+1/2)^2 + (-49/4)}}}
which simplifies to:
{{{y = (x+1/2)^2 + (-49/4)}}}
With h = -1/2 and k = -49/4, the vertex/"turning point" is (-1/2, -49/4).<br>
All that's left is to transform this into standard form. Simplifying...
{{{y = (x)^2+2(x)(1/2)+ (1/2)^2 + (-49/4)}}}
{{{y = x^2+x+ 1/4 + (-49/4)}}}
{{{y = x^2+x+ (-48/4)}}}
{{{y = x^2+x+ (-12)}}}