# Find the quadratic polynomial whose graph goes through the points (-2,10), (0,6) and (1,10).

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A quadratic polynomial has a form `y =ax^2+bx+c` .

To determine the value of a,b and c, plug-in the given points.

For (0,6), plug-in x=0 and y=6.

`6=a*0^2+b*0+c`

`6=c`

Hence, value of c is 6.

Now that value of c is known, plug-in this to the quadratic form of a polynomial.

`y=ax^2+bx+6`

Next, use (-2,10). Substitute x=-2 and y=10 to `y=ax^2+bx+c` .

`10=a(-2)^2+b*(-2)+6`

`10=4a-2+6`

Bring together the constants.

`10-6=4a-2b+6-6`

`4=4a-2b`

Divide both sides by 2, to simplify.

`(4/2)=(4a-2b)/2`

`2=2a-b` (Let this be EQ1.)

Next, use (1,10). Substitute x=1 and y=10 to `y=ax^2+bx+6` .

`10=a*1^2+b*1+6`

`10=a+b+6`

`10-6=a+b+6-6`

4=a+b (Let this be EQ2. )

To solve a and b, apply elimination method of system of equation.

So, add EQ1 and EQ2.

`2=2a-b`

`(+)` `4=a+b`

`------------`

`6=3a+0`

`6=3a`

`6/3=(3a)/3`

`2=a`

Then, plug-in a=2 to EQ1 to solve for b.

`2=2a-b`

`2=2*2-b`

`2=4-b`

`2-4=4-4-b`

`-2=-b`

`(-2)/(-1)=-b`

`2=b`

Now that values of a, b and c are know, plug-in these values to the quadratic form.

`y=ax^2+bx+c`

`y=2x^2+2x+6` **Hence, the quadratic polynomial is `y=2x^2+2x+6` .**