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.
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?
3 answers
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.
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.
3 answers