How do you find slope of the tangent line to the curve #y=x^2+3x-5# at the point (2,5)?

The slope of the tangent line to a curve in a point is the first derivative in that point, so:

and

To find the slope of the tangent line to the curve at the point (2,5), we need to find the derivative of the curve at that point. The derivative of the curve y=x^2+3x-5 is given by the equation dy/dx = 2x + 3. Evaluating this derivative at x=2, we get dy/dx = 2(2) + 3 = 7. Therefore, the slope of the tangent line to the curve at the point (2,5) is 7.

To find the slope of the tangent line to the curve (y = x^2 + 3x - 5) at the point ((2, 5)), you need to find the derivative of the function with respect to (x), and then evaluate it at the given point.

First, find the derivative of the function: [y' = \frac{d}{dx}(x^2 + 3x - 5)] [y' = 2x + 3]

Then, evaluate the derivative at (x = 2): [y' = 2(2) + 3] [y' = 4 + 3] [y' = 7]

So, the slope of the tangent line to the curve at the point ((2, 5)) is (7).