Find the tangent line at $x = -1$

Question

Find the equation of a tangent line for $-3-xy^{2}+3x^{3}=0$ at $x = -1$

I do not even know where to start other than trying to find the deriative

Answer 1

To start, you'll need to find the value of $y$ (or, in this case, values) when $x=-1$ so that you can perform implicit differentiation. Once you do that, you not only will be able to find the slope of the tangent at that point, but you will also have a point on the line with which to create your equation.

Answer 2

This problem requires the use of implicit differentiation. The given implicit relation is:

$-3 - xy^{2} + 3x^{3} = 0$

Differentiating both sides:

$-(y^{2} + 2xyy') + 9x^{2} = 0$

Solving for $y'$:

$y' = \frac{9x^{2}-y^{2}}{2xy}$

Now, we need to find $y$ such that $(-1,y)$ satisfies the relation. We have:

$-3+y^{2}-3 = 0$

$y = \pm\sqrt{6}$

Then, substituting:

$y'\big\vert_{(-1,\sqrt{6})} = -\frac{\sqrt{6}}{4}$

$y'\big\vert_{(-1,-\sqrt{6})} = \frac{\sqrt{6}}{4}$

Thus, the tangent lines are $y-\sqrt{6} = -\frac{\sqrt{6}}{4}(x+1)$ and $y +\sqrt{6} = \frac{\sqrt{6}}{4}(x + 1)$

Simplifying, we obtain $\boxed{y = -\frac{\sqrt{6}}{4}x + \frac{3\sqrt{6}}{4}\text{ and }y =\frac{\sqrt{6}}{4}x - \frac{3\sqrt{6}}{4}.}$

Answer 3

for $x=1$, then, $y=\pm \sqrt6$

$y'=\frac{9x^2-y^2}{2xy}$

$y'\mid_{(-1,\pm \sqrt6)}=\mp\frac{3}{2\sqrt6}$

$y=\mp\frac{3}{2\sqrt6}(x-(-1))+(\pm \sqrt6) $