f(x)=3-5x , (-1,8) find the slope of the tangent line to the graph of the function at the given point
how do you figure this out?
since your function is a straight line
and it is in the form f(x) = mx + b
the slope anywhere on the line is -5
no work needed here.
To find the slope of the tangent line you have to take the first derivative of f(x). So f'(x)=-5. Or you can also see that from the formula y=mx+b, where m is slope and b is your y intercept, that -5 is the slope. Because the original function is a line, the slope is the same at every point.
To find the slope of the tangent line at a point means to find the derivative of f(x).
f(x) = 3 - 5x
f' = 0 - 5
f' = -5
Then, normally, you would plug in the given point, (-1, 8), to find the slope.
But since f' = 5 with no variable x, for instance, there is nowhere to plug in the point.
Therefore, the slope is everywhere.
To find the slope of the tangent line to the graph of the function at the given point, (-1,8), you need to compute the derivative of the function f(x) = 3-5x and then substitute x = -1 into the derivative formula.
Here are the steps to find the slope of the tangent line:
Step 1: Take the derivative of the function f(x) = 3-5x with respect to x.
The derivative of f(x) = 3-5x can be found using the power rule of differentiation. Since the derivative of a constant term is zero, the derivative of 3 is zero. The derivative of -5x can be found using the power rule, which states that d/dx of x^n = nx^(n-1). Therefore, the derivative of -5x is -5.
Step 2: Substitute x = -1 into the derivative.
Plug x = -1 into the derivative from Step 1 to find the slope at that particular point.
So, the slope of the tangent line is the value of the derivative when x = -1, which is -5.
Therefore, the slope of the tangent line to the graph of the function f(x) = 3-5x at the point (-1, 8) is -5.