Question 3 of 20
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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).