# Find the quadratic polynomial `f(x)=ax^2+bx+c` whose graph goes through the points (-2,8), (0,2), and (2,20).

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Since we know that the graph passes through the point `(0,2),` we have `f(0)=c=2,` so we know the constant term and we get`f(x)=ax^2+bx+2.`

Now we plug in another point, say `(-2,8).` This gives

`8=f(-2)=a(-2)^2+b(-2)+2=4a-2b+2.` The last point gives

`20=f(2)=a(2)^2+b(2)+2=4a+2b+2,` so we end up with the system of equations

`4a-2b=6`

`4a+2b=18.`

If we add them, theĀ `b`'s cancel and we get `8a=24,` so `a=3.` Now we can plug this value of `a` into the top equation and get

`4(3)-2b=6,` so `b=3.`

**The quadratic polynomial that goes through those points is `f(x)=3x^2+3x+2,` which can be easily checked by plugging in.**