Find the equation of the tangent line to the graph of f(x) = x^3 such that the tangent line equation has an x-intercept of x = 2 and isn't the x-axis.

Answer 1

The tangent line at any point on the graph where the POT (point of tangency) has an x-coordinate a, is given by::

y=3a²(x-a) + a³.

This is because the formula for the slope of any tangent line is y' = 3x² and when x = a, y = a³, i.e. a point the line passes through is (a, a³), so using point-slope yields the given equation.

We want the particular equation that passes through the point (2,0), so substitute 2 for x and 0 for y in that equation and solve for a.

Confirm here:

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Answer 2

The tangent line at any point "x" has a slope that equals the derivative at "x".

If f(x) = x³, then f '(x) = 3x²

The point-slope form of a line is: y - y₁ = m(x - x₁) where "m" is the slope. We now the slope is 3x². We also know (since the x-intercept is x = 2) that a point on the line is (2, 0) so we can plug those into the point-slope form:

y - 0 = 3x²(x - 2)

y = 3x³ - 6x²

Now, consider the point of tangency, We don't know what it is but we can say that when x = x, y will be x³ meaning another point on our tangent line is (x, x³) and we can plug that in to get:

x³ = 3x³ - 6x²

0 = 2x³ - 6x²

0 = 2x²(x - 3)

x = 0 or x = 3

x = 0 means y =0 which will give us a tangent line of the x-axis so we can throw that out.

If x = 3, and m = 3x², then m = 27 so the point-slope form equation becomes:

y - 0 = 27(x - 2)

y = 27x - 54

Answer 3

let's try

want slope f'(x₀)=[f(x₀)-0]/[x₀-2],,,,,line will connect points (2,0) and (x₀,f(x₀))

or dropping subscript, 3x²=x³/(x-2)

x=3

now 27=[y-0]/[x-2] for equation

or y=27x - 54.