# I need to find the value of n if the graph of f(x)=n*x+n-4 is passing through the point A(n,-2).

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We need to find n if f(x) = n*x + n - 4 passes through the point A(n, -2)

Substituting f(x) = -2 and x = n gives

-2 = n^2 + n - 4

=> n^2 + n - 2 = 0

=> n^2 + 2n - n - 2 = 0

=> n(n + 2) - 1(n + 2) = 0

=> (n - 1)(n + 2) = 0

=> n = 1 and n = -2

**The values of n are n = 1 and n = -2**

If the point A lies on the graph of the given function, it's coordinates must verify the equation of the function.

A(n,-2) belongs to graph of f(x) <=> f(n) = n*n + n - 4 = -2

We'll compute f(n):

n^2 + n - 2 = 0

We'll apply quadratic formula:

n1 = [-1 + sqrt(1 + 8)]/2

n1 = (-1 + 3)/2

n1 = 1

n2 = (-1 - 3)/2

n2 = -2

Since there are no conditions imposed concerning the nature of the value of n, we'll keep them both, therefore there are two points that may be located on the graph of f(x): (1 ; -2) and (-2 , -2).

**The requested values for n are: {-2 ; 1}.**