Find a point on the curve y = 3x ^ 2-5x-11 whose tangent is parallel to the line x-y + 10 = 0.

Let us denote the abscissa of the required point by a. First, we compose the equation of the tangent to the graph of the function y = y (x) = 3 * x ^ 2 – 5 * x – 11 at the point a, for which we use the equation of the tangent y = y (x0) + yꞌ (x0) * (x – x0 ) to the graph of the function y = y (x) at the point x0.
Find the derivative yꞌ = (3 * x ^ 2 – 5 * x – 11) ꞌ. We use the properties of differentiation: (u ± v) ꞌ = uꞌ ± vꞌ, Cꞌ = 0, where C is a constant, as well as a table of derivatives of basic elementary functions. We have yꞌ = yꞌ (x) = (3 * x ^ 2) ꞌ – (5 * x) ꞌ – 11ꞌ = 6 * x – 5.
We find y (a) = 3 * a ^ 2 – 5 * a – 11 and yꞌ (a) = 6 * a – 5.
So, the equation of the tangent y = 3 * a ^ 2 – 5 * a – 11 + (6 * a – 5) * (x – a) or y = (6 * a – 5) * x – 3 * a ^ 2 – eleven.
Now we use the condition of parallelism of two lines y = k1 * x + b1 and y = k ^ 2 * x + b ^ 2. It is very simple to formalize k1 = k2. This equation x – y + 10 = 0 is reduced to the form y = k * x + b. We have y = x + 10. The tangent equation has the form y = (6 * a – 5) * x – 3 * a ^ 2 – 11.
Therefore, 6 * a – 5 = 1 or 6 * a = 6, whence a = 1. Now we define the ordinate of the point of contact. It is equal to y (a) = 3 * 1 ^ 2 – 5 * 1 – 11 = 3 – 5 – 11 = –13. Thus, the required point has coordinates: (1; –13).
Answer: (1; –13).