f(x)=3-5x , (-1,8) find the slope of the tangent line to the graph of the function at the given point
f' = -5 = slope everywhere
how did you figure that out?
To find the slope of the tangent line to the graph of the function at the given point, we can use the derivative of the function. The derivative represents the rate at which the function is changing at any given point.
Let's find the derivative of the function f(x) = 3 - 5x first:
f'(x) = d/dx (3 - 5x)
To find the derivative of a constant term (3) with respect to x, we get 0 because the derivative of a constant is always 0.
Thus, we only need to differentiate -5x with respect to x. To do this, we can use the power rule of differentiation, which states that if we have a term of the form ax^n, then the derivative is given by n * ax^(n-1).
In this case, we have -5x, where the coefficient is -5 and the exponent is 1. Applying the power rule, we get:
f'(x) = 0 - 5 * 1 * x^(1-1)
= -5
Now that we have the derivative, we can find the slope of the tangent line at the given point (-1, 8). The slope of the tangent line is equal to the derivative of the function evaluated at that point.
So, the slope of the tangent line is f'(-1) = -5.