# What is the quadratic function that is created with roots -10 and -6 and a vertex at (-8,-8)?

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### 1 Answer

You need to write the quadratic polynomial in factored form such that:

`f(x) = a(x-x_1)(x-x_2)`

`x_1,x_2` denote the roots of quadratic

You may write as well the quadratic in vertex form such that:

`f(x) = a(x-h)^2 + k^2`

h,k denotes coordinates of vertex of parabola

Plugging the roots in the factored form yields:

`f(x) =a(x-(-10))(x-(-6))`

`f(x) =a(x+10)(x+6) =gt f(x) =a(x^2 + 6x + 10x + 60)`

Adding like powers yields:

`f(x) =a(x^2 + 16x+ 60)`

Plugging the coordinates of vertex in the vertex form yields:

`f(x) = a(x+8)^2 + (-8)^2`

Expanding the binomial yields:

`f(x) = a(x^2 + 16x + 64)+ 64`

Equating the two forms yields:

`a(x^2 + 16x+ 64) + 64= a(x^2 + 16x + 60)`

Move the terms containing the coefficient a to the left side such that:

`a(x^2 + 16x+ 64)- a(x^2 + 16x + 60)= -64`

Factoring out a yields:

`a(x^2 + 16x + 64 - x^2 - 16x - 60) = -64`

Reducing like terms yields:

`4a = -64 =gt a = -16`

`` **Hence, evaluating the quadratic function yields`f(x)=-16x^2 - 256x - 960.` **