To write a quadratic equation in intercept form with x-intercepts at 3 and 6, we can start with:
(x - 3)(x - 6) = 0
Expanding this equation gives:
x^2 - 9x + 18 = 0
Now we need to use the point (2, -2) to find the equation in intercept form.
First, substitute x = 2 and y = -2 into the standard form of a quadratic equation:
y = ax^2 + bx + c
-2 = a(2)^2 + b(2) + c
-2 = 4a + 2b + c
Next, substitute the x-intercepts into the equation:
y = a(x-3)(x-6)
We want the equation in intercept form, so we expand this equation:
y = a(x^2 - 9x + 18)
y = ax^2 - 9ax + 18a
Now we can use the three points we have to form a system of equations:
-2 = 4a + 2b + c
0 = 9a - 54a + 18a
0 = 3a
The last equation tells us that a = 0, which means the quadratic equation is a linear equation:
y = bx + c
To solve for b and c, we use the other two equations:
-2 = 4a + 2b + c
0 = 9a - 54a + 18a
Plugging in a = 0 into the first equation, we get:
-2 = 2b + c
Simplifying the second equation gives:
0 = -27a
Since a = 0, this equation is true and does not give us any new information.
We can now solve for b and c by solving the system of equations:
-2 = 2b + c
0 = -27a
Since a = 0, we can ignore the second equation and solve for b and c using the first equation:
2b + c = -2
We still have one degree of freedom, so we can choose any value for b and solve for c. Let's choose b = 1:
2(1) + c = -2
c = -4
Therefore, the equation of the quadratic function in intercept form is:
y = (x-3)(x-6)
or
y = x^2 - 9x + 18
which has x-intercepts at 3 and 6 and passes through the point (2, -2).