find an equation for the line tangent to the curve at the given point and show how
y=(x^4+2)/x^2, x=-1
y = x^2 + 2/x^2
y(-1) = 3
y' = 2x - 4/x^3
y'(-1) = -2 + 4 = 2
Now we have a slope=2 and a point (-1,3).
y-3 = 2(x+1)
y=(x^4+2)/x^2, x=-1
y(-1) = 3
y' = 2x - 4/x^3
y'(-1) = -2 + 4 = 2
Now we have a slope=2 and a point (-1,3).
y-3 = 2(x+1)