Question 503419
Looking at the graph, you can see that y = 4/x is not truly a linear equation.
{{{graph(500,500,-10,10,-10,10,4/x)}}}
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Actually
{{{y = 4 * x^(-1)}}}
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The point (-2,8) is not on the curve, which you see above.  Or prove it by plugging the x value into the equation:
y = 4/x
x = -2
y = -2, not 8

That creates another problem for you.
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In general the slope of a line tangent to a curve can be found using the first derivative.
But that is a calculus question, not algebra.
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The first derivative of 
{{{4/x = -4/x^2}}}
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That is the slope of the line tangent to the curve for any value of 'x'.
But you need a point on the curve, which you do not have.
So, there is something wrong with your problem statement.
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But assuming the point = (-2,-2)
The slope is:
-4/(-2^2) = -4/4 = -1 at that point.
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Remember this is a curve, so the slope changes continuously.
(except there is a discontinuity at x=0 because 4/0 is undefined)
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Substituting in the slope-intercept equation:
y = mx+b

-2 = -1*(-2) + b
b = -4
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y = -x -4
which is a linear equation.
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{{{graph(500,500,-10,10,-10,10,4/x,-x-4)}}}
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Of course, you could have picked another point.
Let's say x =4
The curve is y = x/4, so y=1.
(4,1)
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m = -4/(4^2) = -4/16 = -1/4
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1 = -1/4(4) + b
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so
b=2
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y = (-1/4)x+2
which is a linear equation tangent to the curve at (4,1)

{{{graph(500,500,-10,10,-10,10,4/x,-x-4,-1/4*x+2)}}}