# tangent line equation

• Apr 10th 2008, 01:01 PM
Eric08
tangent line equation
Hey, im having some problems with this

Find the equation of the line tangent to the curve y=2x^4-x^3 at the point (-1, 3). Any help appreciated.

i plugged the ordered pair into the genreal equation of a line, and now im stuck. What would the dy/dx for y=2x^4-x^3 be?

any help appreciated, thanks
• Apr 10th 2008, 01:07 PM
Kalter Tod
You just wanna find the derivative of the original function...

So if \$\displaystyle y=2x^4-x^3\$ then \$\displaystyle y'=8x^3-3x^2\$

Use the x value you were given in the point (-1,3) and you will get that

\$\displaystyle y'(-1)=-11\$

So, now that you have a slope and a point, you can find the slop of the line by using the formula \$\displaystyle (y-a)=m(x-b)\$ So you get the equation of the line to be \$\displaystyle y=-11x+32\$
• Apr 10th 2008, 01:11 PM
Eric08
wait, shouldnt the final answer be y=-11x-8
• Apr 10th 2008, 01:17 PM
Kalter Tod
Actually...on second thought...I mixed up my numbers a little bit there

It should be \$\displaystyle y=-11x-14\$
• Apr 10th 2008, 01:25 PM
Eric08
Quote:

Originally Posted by Kalter Tod
Actually...on second thought...I mixed up my numbers a little bit there

It should be \$\displaystyle y=-11x-14\$

when i plug this into my calculator it doesnt even touch the parabola. isnt the answer supposed to be y=-11x-8, thats what i got when I solved and thats what the calc is saying for a tangent line to the parabola at X=-1
• Apr 10th 2008, 01:27 PM
Kalter Tod
Haha yeah it is...Apparently, I can't add numbers.

I did it all in my head, so sorry for the math mistakes haha
• Apr 10th 2008, 01:29 PM
Eric08
no problem