# Determine the quadratic function f(x)= ax^2+bx+c whose vertex is (-2,-11) and passes through the point (-5,7). Please show all work, thanks.

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You need to remember that the vertex of quadratic function denotes the critical point of the function such that: f'(-2) = 0.

Hence you need to differentiate the function with respect to x and then you need to substitute -2 for x in equation of derivative such that:

`f'(x) = 2ax + b =gt f'(-2) = 0 =gt -4a + b = 0 =gt 4a = b`

You need to remember that the coordinates of a point on a graph check the equation of the representing function.

The vertex is a point on the graph, hence substituting -2 for x in equation of function yields -11 such that:

`f(-2)= a(-2)^2+b*(-2)+c =gt -11 = 4a - 2b + c`

You need to substitute b for 4a in `4a - 2b + c=-11 =gt b-2b+c=-11`

`-b+c = -11 =gt c = b - 11 =gt c = 4a -11`

The point (-5,7) is on the graph, hence substituting -5 for x in equation of function yields 7 such that:

`f(-5) = 25a - 5b + c=gt 25a - 5b + c = 7`

You need to write the equation `25a - 5b + c = 7` in terms of a such that:

`25a - 5*4a + 4a - 11 = 7 =gt 9a = 11 + 7 =gt a = 18/9=gt a=2`

Substituting 2 for a in b = 4a yields: b = 8.

Substituting 2 for a in `c = 4a -11 =gt c = 8 - 11 =gt c = -3`

**Hence, evaluating the coefficients a,b,c yields: `f(x)=2x^2 + 8x - 3.` **

Given the vertex of a parabola at (-2,-11) and that the parabola goes through the point (-5,7).

The vertex form of a parabola is `y=a(x-h)^2+k` where the vertex is at the point `(h,k)` .

So for the vertex (-2,-11) we have `y=a(x+2)^2-11` .

In order to find `a` we plug in the known point (-5,7):

`7=a(-5+2)^2-11`

`7=9a-11`

`a=2`

**Thus the equation is `y=2(x+2)^2-11` or in standard form **`y=2x^2+8x-3`