# What is the equation of the quadratic function with roots -2 and 4 and a vertex at (1,9)

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If the quadratic function has roots -2 and 4, we can write f(x) = (x - (-2))(x - 4)

=> f(x) = (x + 2)(x - 4)

=> f(x) = x^2 - 4x + 2x - 8

=> f(x) = x^2 - 2x - 8

The vertex of the function is the minimum point. This is given at the point where f'(x) = 0

=> 2x - 2 = 0

=> x = 1

f(1) = 1 - 2 - 8 = -9

The vertex for the function is (1, -9).

For the given roots the quadratic function that we arrive at has the vertex at (1, -9), not (1,9)

**The quadratic function with roots -2 and 4 is f(x) = x^2 - 2x - 8.**

Let the function be f(x).

Since -2 and 4 are the roots for the quadratic function, then the factors are ( x+ 2) and (x-4)

==> f(x) = (x+2)(x-4)

==> f(x) = x^2 -2x - 8

Let us determine the vertex.

The vertex is given by V(vx, vy) such that:

==> vx = -b/2a = 2/2 = 1

==> vy = (b^2-4ac)/4a = ( 36)/4 = 9

Then the vertex is (1,9)

**Then the equation of the function is f(x) = x^2 -2x-8 **