hi guys im having trouble figuring this out.
Find the equation of the tangent to the curve
$\displaystyle
y = \frac {x^2} {x-1}
$
when
$\displaystyle
x = -1
$
thank u!
Hello
Let $\displaystyle f(x)=\frac {x^2} {x-1}$
The equation of the tangent to the curve of $\displaystyle f$ at $\displaystyle a$ is $\displaystyle y=f(a)+(x-a)f'(a)$. In this case, $\displaystyle a=-1$ hence what you've to do is simply computing $\displaystyle f(-1)$ and $\displaystyle f'(-1)$.
To find the derivative of $\displaystyle f$, you can use the quotient rule : $\displaystyle \left(\frac{u}{v}\right)'=\frac{u'v-v'u}{v^2}$
Does it help ?
yes we did that but what we are confused about is if we use x = -1 for both points?
because we will input -1 into the original formula, then the first point will be (-1, y(original))
then we get the derivative of the original equation then input x into that equation which will give us (-1,y(derivative of original)) ??????
but then if we find m, the denominator will be equal to ZERO which will give us an undefined function.
??
You find the equation of the tangent by using $\displaystyle y-y_{1}=m(x-x_{1})$ where $\displaystyle x_{1}$ and $\displaystyle y_{1}$ are coordinates for a point on the tangent, in your case, $\displaystyle (-1,f(a))$. m is found by substituting -1 into $\displaystyle y'$, which you get by using the quotient rule.