# Thread: Tangent

1. ## Tangent

find the equation of a line that is tangent to the graph of the given function at the specified point.
y=x^5-3x^3-5x+2;(1,-5)

I have an exam coming up on these and I pulled this out of my book to practice...if you could show how to do this one so I can try the rest on my own it would be a big help..Thanks!

2. Originally Posted by lisa1984wilson
find the equation of a line that is tangent to the graph of the given function at the specified point.
y=x^5-3x^3-5x+2;(1,-5)

I have an exam coming up on these and I pulled this out of my book to practice...if you could show how to do this one so I can try the rest on my own it would be a big help..Thanks!

The derivative function will be used to measure slope at the given point. Once we have the slope we can use the point-slope formula to find the equation of the tangent line.

The derivative is $f'(x) = 5x^4 - 9x^2 - 5$

Evaluation at $x = 1$, we get $f'(1) = 5 - 9 - 5 = -9$

so slope is -9 and passes through (1, -5).

Good luck!

3. ## Tangent

y=2x^4- √x+3/x;(1,4)
derivative I got is:
8x^3-1/2-3/x^2
which then I solved by filling in the x with 1 and got the answer 4.5 slope

4. Originally Posted by lisa1984wilson
y=2x^4- √x+3/x;(1,4)
derivative I got is:
8x^3-1/2-3/x^2
which then I solved by filling in the x with 1 and got the answer 4.5 slope

Lisa:

your derivative seems incorrect. The derivative of $\sqrt{x}$ is $\frac{1}{2} x^{\frac{-1}{2}}$.

5. radicals are not my thing but I need to get them down...
I got this as the derivative:
8x^3-11/4x^2

6. The derivative is:

$8x^3-\frac{1}{2\sqrt{x}} -\frac{3}{x^2}$