Find n if the area of the triangle determined by the graph of the function f=nx+n-4 and the x,y axis where is A=n/2.

The area of the triangle whose sides are the graph of f(x)
and x, y axis is the area of a right angle triangle: half-product of
cathetus.

The cathetus of the triangle are x and y axis,
where the values of x and y are the intercepts of the graph with x and y
axis.

A = x*y/2

We'll
calculate x intercept of f(x). We'll put y = 0.

f(x) =
0

nx+n-4 = 0

We'll isolate x
to the left side:

nx = 4 - n

x
= (4-n)/n

We'll calculate y intercept of f(x). We'll put x
= 0.

f(0) = n - 4

y = n -
4

Now, we'll calculate the
area:

A = (4-n)*(n-4)/2n

We
know, from enunciation, that A=n/2.

n/2 =
-(n-4)^2/2n

We'll cross
multiply:

2n^2 =
-2(n-4)^2

We'll divide by
2:

n^2 = (n-4)^2

We'll
subtract (n-4)^2 both sides:

n^2 - (n-4)^2 =
0

We'll expand the
square:

n^2 - n^2 + 8n - 16  =
0

We'll eliminate like
terms:

8n - 16 = 0

We'll add
16

8n = 16

n =
2