# What is the angle of the tangent to the graph of the function y = x ^ 2 + 4x-5 drawn at the point of the graph

**What is the angle of the tangent to the graph of the function y = x ^ 2 + 4x-5 drawn at the point of the graph with the abscissa x0 = 1 with the positive direction of the x axis.**

First, we compose the equation of the tangent to the graph of the function y = y (x) = x² + 4 * x – 5 at the point x0 = 1, 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.

Let’s find the derivative of this function. We use the formulas: (u ± v) ꞌ = uꞌ ± vꞌ, (C * u) ꞌ = C * uꞌ, Cꞌ = 0, (un) ꞌ = n * un – 1 * uꞌ, where C and n are constants. We have: yꞌ (x) = (x² + 4 * x – 5) ꞌ = (x²) ꞌ + (4 * x) ꞌ – 5ꞌ = 2 * x + 4 * xꞌ – 0 = 2 * x + 4 * 1 = 2 * x + 4.

We calculate y (x0) = y (1) = 1² + 4 * 1 – 5 = 1 + 4 – 5 = 0 and yꞌ (x0) = yꞌ (1) = 2 * 1 + 4 = 6.

So, the equation of the tangent has the form: y = 0 + 6 * (x – 1) or y = 6 * x – 6.

The derivative at the point of tangency is 6, which means that the tangent of the slope is 6. Find the desired angle, which we denote by α, using the equality tan (α) = 6. So, α = arctan (6).

Answer: arctg (6).