# A quadratic equation is a parabola that has the roots x = 3 and x = -5. It passes through the point (1, -12). Determine its quadratic equation.

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A quadratic equation has a form:

`y=ax^2 + bx + c`

To determine the values of a, b and c, plug-in the given roots. Note that roots are the values of x when y is zero.

So, when we plug-in the root x=3, the resulting equation is:

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

`0=9a+3b+c` (Let this be EQ1.)

And, when we plug-in the roots x=-5, the resulting equation is:

`0=a(-5)^2+b(-5)+c`

`0=25a-5b+c` (Let this be EQ2.)

Next, plug-in the point (1, -12).

`-12=a(1)^2+b(1)+c`

`-12=a+b+c` (Let this be EQ3.)

Now that we have three equations, to solve for values of a, b and c, apply elimination method.

To do so, subtract EQ3 from EQ2 to eliminate the c.

`0=9a+3b+c`

`(-)` `-12=a+b+c`

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

`12=8a+2b`

`6=4a+b` (Let this be EQ4.)

Next, subtract EQ3 from EQ2 to eliminate c again.

`0=25a-5b+c`

`(-)` `-12=a+b+c`

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

`12=24a-6b`

`2=4a-b` (Let this be EQ5.)

Then, add EQ4 and EQ5 to eliminate b.

`6=4a+b`

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

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

`8=8a`

`1=a`

Plug-in this value of a to either EQ4 or EQ5 to solve for b.

So, using EQ4, the value of b is:

`6=4a+b`

`6=4(1)+b`

`6=4+b`

`2=b`

Then, plug-in values of a and b to either EQ1, EQ2 or EQ3.

Using EQ3, the value of c is:

`-12=a+b+c`

`-12=1+2+c`

`-12=3+c`

`-15=c`

Now that the values of a, b and c are known, substitute their values to the form of quadratic equation which is y =ax^2+bx+c.

`y=(1)x^2+(2)b+(-15)`

`y=x^2+2x-15`

**Hence, the quadratic equation that has roots x=3 and x=-5 and passes the point (1,-12) is `y=x^2+2x-15` .**