Find the coinciding point(s) of all the tangent(s) to the curve y=3x^2-2 that pass through the point(1,-11).

Answer 1

The function and the tangent line both share the same point (1, -11) or the tangent line would not be tangent. it would be just some line on the graph that may or may not intersect the at all. Because the function is parabolic, any tangent line would intersect only one time.

If f(x) = 3x^2 -2, then f'(x) = 6x (by power rule). This is the derivative. More specifically, however, at x = 1, the m of the line is 6x or 6. You know have a point and a slope. Using y=mx+b

-11 = 6(1) + b

b = -17

Therefore the equation of the tangent line is y = 6x-17

Answer 2

Let (x₀, y₀) is tangent point. Then y₀ = 3x₀² - 2 and slope of tangent line is y'(x₀) = 6x_0;

Equation of tangent line y = y₀ + 6x₀(x - x₀). And this tangent line has a point (1, - 11). So,

-11 = y₀ + 6x₀(1 - x₀). Now we have system of 2 equations for x₀ and y₀:

y₀ = 3x₀² - 2 and y₀ = 6x₀² - 6x₀ - 11;

3x₀² - 2 = 6x₀² - 6x₀ - 11; 3x₀² - 6x₀ - 9 = 0; x₀² - 2x₀ - 3 = 0; (x₀ - 3)(x₀ + 1) = 0; x₀ = 3 or x₀ = - 1;

y₀ = 3·3² - 2 = 25 or y₀ = 3(- 1)² - 2 = 1.

We get 2 points: (3, 25) and (- 1, 1)