# find the general equation of the parabola that passes through the three points (1,3), (-1,9) and (2,6)

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### 1 Answer

A parabola is an equation of the form `y=ax^2+bx+c`

(where `a != 0` )

If we had a,b, and c, we would have the equation.

We know the parabola passes through (1,3), which means we can plug in 1 for x and 3 for y, and the equation should be satisfied. So:

`3=a(1)^2+b(1)+c=a+b+c`

Similarly for (-1,9) and (2,6)

So we have:

`9=a(-1)^2+(-1)b+c`

`6=a(2)^2+(2)b+c`

Thus:` `

`3=a+b+c`

`9=a-b+c`

`6=4a+2b+c`

We want to solve for a,b,c. Subtract the first equation from the second equation:

`6=-2b` so `b=-3`

Plug this into the equations to get:

`3=a-3+c`

`9=a+3+c`

`6=4a-6+c`

The first two equations essentially say the same thing. Simplifying, we have:

`6=a+c`

`12=4a+c`

Subtract the first from the second, to get:

`6=3a` so `a=2`

Plug this into either equation:

`6=2+c`

We must have `c=4`

So:

`a=2,b=-3,c=4`

The equation must be `y=2x^2-3x+4`

(If you plot that graph, you can see that it really does go through the points)

**The parabola that passes through the points is:**

`y=2x^2-3x+4 `