Answer:
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Step-by-step explanation:
We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).
Remember that the standard quadratic function is given by:
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Since it passes through the point (-3, 2). This means that when
,
. Hence:

Simplify:

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

And for (1, 6):

So, we have a triple system of equations:
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We can solve this using elimination.
Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

Compute:
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Let's divide both sides by 2:

Now, let's eliminate b again but we will use Equation 1 and 2.
Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

Multiply:

Add this to Equation 1:

Compute:

Again, we can divide both sides by 2:

So, we know have two equations with only two variables:

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

Compute:

Solve for a:

So, the value of a is 1.
Using either of the two equations, we can now find c. Let's use the first one. Hence:

Substitute 1 for a and solve for c:

So, the value of c is 2.
Finally, using any of the three original equations, solve for b:
We can use Equation 3. Hence:

Substitute in known values and solve for b:

Therefore, a=1, b=3, and c=2.
Hence, our quadratic function is:
