The tangent to the curve y = x−3x^(-1) at (x, y) = (1,−2) meets the x-axis at (a,0) where a =?

y = x - 3x^(-1)

dy/dx = 1 + 3/(x^2)

Slope of tangent to curve at (1, - 2) = 1 + 3 = 4

Eqn of tangent line:

y + 2 = 4(x - 1)

y = 4x - 6

Find the x-intercept:

4x - 6 = 0

x = 3/2

Therefore, a = 3/2

it means the line given by the tangent to the curve at (1,-2) intersects the x axis at a certain point (a,0)

so basically you have to find the equation of a line where the slope (m) is equal to y' at x= 1.

so... y = m*x +b

once you have m (just sub in x = 1 into the derivative of y (y')) you can set x and y to 1 and -2 in your y = mx+b equation and then solve for b.

Now that you have the equation of the line, check what x is equal to when you set y = 0;

this will give you the a value