A quadratic function has a vertex (3, -6) and the point (-1, 10). Write the function in its general form.

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The graph of a quadratic function is a parabola.

The standard (or vertex) form is `f(x) = a(x-h)^2+k`

where h and k are the co-ordinates of the vertex (3;-6).

The general form is `f(x)=ax^2+bx+c`

As we have the vertex (3;-6) substitute into that equation:

`f(x) = a(x-3)^2 +(-6)` . Note that as x=3, the factor is x-3.

`thereforef(x)= a(x-3)^2-6`

We also have the point (-1;10) where x=-1 and y or f(x)=10

So to find a, we can say `f(-1) =10`

Substituting `f(x) = a(x-3)^2-6` becomes

`10=a(-1-3)^2 -6`

`therefore 10= a(-4)^2-6`

`therefore 10=16a-6`

`therefore 10+6=16a`

`therefore 16/16=a`

`therefore a=1`

So in standard (vertex) form : `f(x)=1(x-3)^2-6`

To convert to general form, expand the brackets:

`f(x)=x^2 -6x +9 -6`

`therefore f(x) = x^2-6x+3`

**Ans: f(x) = x^2-6x+3**

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