# Thread: tangent to the function #2

1. ## tangent to the function #2

1. Find the values of x so that the tangent to the function y=3/(x^1/3) is parallel to the line x+16y+3=0?

2. Find the point on the parabola y=-x^2+3x+4 where the slope of the tangent is 5

2. 1. The slope of the tangent line to the graph of a function at a point x is given by the derivative of the function at that point. Thus, the optimal approach is to find the derivative of the first function, and find where it is equal to the slope of the given line.

2. This is the same, except you are now equating the derivative to the value 5 and solving for x.

Show some work and show where you are having difficulty.

3. Originally Posted by slider142
1. The slope of the tangent line to the graph of a function at a point x is given by the derivative of the function at that point. Thus, the optimal approach is to find the derivative of the first function, and find where it is equal to the slope of the given line.

2. This is the same, except you are now equating the derivative to the value 5 and solving for x.

Show some work and show where you are having difficulty.

OK>>>>>
1.
y'=-x^(-4/3)
now... i dont know what to do with the derivative.

2. y'=-2x+3
but still don't know what to do

thank you

4. Originally Posted by skeske1234

2. y'=-2x+3
but still don't know what to do

thank you
Make y' = 5 and solve for x.

5. Originally Posted by skeske1234
OK>>>>>
1.
y'=-x^(-4/3)
now... i dont know what to do with the derivative.
The derivative is the slope of the tangent line to the graph of y = f(x) at the point x. You thus want to set the derivative y' = -x^(-4/3) equal to the slope of the line given by the next equation and find the value of x satisfying that equality.

2. y'=-2x+3
but still don't know what to do

thank you
As before, you now have the slope of the tangent line to the graph at the point x, which is -2x + 3. The question asks for which x is that value (y') equivalent to 5.