Line y = 4x + 13 is parallel to the tangent to the graph of the function y = x ^ 2 – 3x + 5.

Line y = 4x + 13 is parallel to the tangent to the graph of the function y = x ^ 2 – 3x + 5. Find the abscissa of the tangent point.

The equation of the tangent to the graph of the function f (x) at the point x = x0 has the following form:

y = f ‘(x0) * (x – x0) + f (x0).

We write the equation of the tangent to the graph of the function f (x) = x ^ 2 – 3x + 5 at the point x = x0.

Find the derivative f ‘(x):

f ‘(x) = (x ^ 2 – 3x + 5)’ = 2x – 3.

Therefore, the equation of the tangent to the graph of the function f (x) = x ^ 2 – 3x + 5 at the point x = x0 has the following form:

y = (2×0 – 3) * (x – x0) + x0 ^ 2 – 3×0 + 5.

According to the problem statement, this tangent is parallel to the line y = 4x + 13.

Therefore, the slopes of these lines must be equal:

2×0 – 3 = 4.

Solving the resulting equation, we find the abscissa of the point of tangency:

2×0 = 3 + 4;

2×0 = 7;

x0 = 7/2 = 3.5.

Answer: 3.5.