# Thread: Find Equation of Tangent Line

1. ## Find Equation of Tangent Line

Find the equation of a tangent line at the given value (in form Ax+Bx+C=0)

$y^3+xy-y=8x^4$ ; x=1

2. first you have to find the slope of the function by finding the derivative which with this function you have to do implictly then plugging in 1and whatever why value we get by plugging 1 into the original function

so
$3y^{2}\frac{dy}{dx}+(y+x\frac{dy}{dx})-\frac{dy}{dx}=32x^3$

now group our dy/dx terms to the left and non dy\dx terms to the right

$3y^2\frac{dy}{dx}+x\frac{dy}{dx}-\frac{dy}{dx}=32x^3-y$

factor out dy/dx

$\frac{dy}{dx}(3y^2+x-1)=32x^3-y$

$\frac{dy}{dx}=\frac{32x^3-y}{3y^2+x-1}$

now plug in your x and y values to get the slope and then use point slope form to get the linear equation for the tangent line

3. I plug in the x=1 and y=o but the slope i get isn't what the answer should be. The answer in the back of the book says it should be $y=\frac{5}{2}x-\frac{1}{2}$ The math you did though is what i got too. Can anyone find out how to get this answer?

Thanks!

4. Still stuck on this, its probably an addition error but anyone who can help?

5. Originally Posted by Jim Marnell
I plug in the x=1 and y=o but the slope i get isn't what the answer should be. The answer in the back of the book says it should be $y=\frac{5}{2}x-\frac{1}{2}$ The math you did though is what i got too. Can anyone find out how to get this answer?

Thanks!
Why are you taking y= 0? At x= 1, your equation. $y^3+xy-y=8x^4$ becomes $y^3+y- y= y^3= 2$. y= 0 does not satisfy that.

6. Originally Posted by HallsofIvy
Why are you taking y= 0? At x= 1, your equation. $y^3+xy-y=8x^4$ becomes $y^3+y- y= y^3= 2$. y= 0 does not satisfy that.
oo ok, so i had 2 plug x=1 into the original equation to get y. But if i plug 1 into 8x^4 wouldnt i get 8 so it would be y^3=8 which gives us y=2