Question 217107
The key to this problem is to understand the symmetry of the parabola. The line of symmetry is x = 4. So the parabola to the left of x = 4 is a mirror image of the parabola to the right of x = 4.<br>
So if there is an x-intercept at 2 (which is 2 to the left of 4), there will be another x-intercept at 6 (2 to the right of 4). So we have two x-intercepts: (2,0) and (6,0). We are also told that the vertex is (4, 6).<br>
Now how do we get the equation? Here are a couple of ways:<ol><li>Since the y-coordinates of the x-intercepts are zero we are looking for an {{{ax^2 +bx +c}}} that will be equal to zero when x = 2 and when x = 6. The simplest way to find this is to use:
{{{y = a(x-2)(x-6)}}}. With some thought you should understand that if x=2 or x=6 then y = 0. Next we will use the other point, the vertex, to find the "a". Se will substitute 4 for the x and 6 for the y:
{{{6 = a(4-2)(4-6)}}} and solve for a:
{{{6 = a(2)(-2)}}}
{{{6 = -4a}}}
{{{-3/2 = a}}}
So our equation is: y = {{{(-3/2)(x-2)(x-6)}}}. If we want it in {{{y = ax^2 +bx +c}}} form then we need to multiply out the right side giving:
{{{y = (-3/2)x^2 + 12x - 18}}}</li><li>Another way to do this is to substitute the three points, one point at a time, into {{{y=ax^2+bx+c}}} giving three equations:
{{{(0) = a*(2)^2 + b(2) + c}}}
{{{(0) = a*(6)^2 + b(6) + c}}}
{{{(6) = a*(4)^2 + b(4) + c}}}
Simplifying each we get:
{{{0 = 4a + 2b + c}}}
{{{0 = 35a + 6b + c}}}
{{{6 = 16a + 4b + c}}}
We now have a system of three equations with three variables (a, b and c). There are a number of methods for solving systems like this, including Substitution, Linear Combination (aka Elimination or Addition), Kramer's rule (determinants) and a variety of matrix methods. Choose the one with which one you are most comfortable. It should result in the same a, b and c we found earlier.</li></ol>