Question

Tangent Line to a Function

- The graph of the function f(x)=sin(x+8) is shown along with the points (-8, f(-8)) and (-7, f(-7)). Draw the tangent line to the point (-8, f(-8)) and plot the point on the tangent

Answer 1

To find the tangent line to the point (-8, f(-8)) on the graph of f(x) = sin(x+8), we need to find the derivative of the function at x = -8. The slope of the tangent line is 1, and the equation of the tangent line is y = x + (f(-8) - 8).

Step-by-step explanation:

To find the tangent line to the point (-8, f(-8)) on the graph of f(x) = sin(x+8), we need to find the derivative of the function at x = -8. The derivative of f(x) = sin(x+8) is f'(x) = cos(x+8).

Substituting x = -8 into f'(x), we get f'(-8) = cos(-8+8) = cos(0) = 1.

So, the slope of the tangent line to the point (-8, f(-8)) is 1. The equation of a straight line with a slope of 1 passing through the point (-8, f(-8)) can be written as y - f(-8) = 1(x - (-8)). Simplifying this equation, we get y = x + (f(-8) - 8). Now we can plot the point (-8, f(-8)) and draw the tangent line.