The line tangent to $y = -x^3 + 2x + 1$ when $x = 1$ intersects the curve in another point. Find the coordinates of the other point.
This was never taught in class, and I have a test on this tomorrow. This question came off of my test review worksheet, and I don't understand how to solve it. The answers are on the back, and for this one it says the answer is (-2,5), but I don't understand how to get that.
I did the derivative and substituted 1 for x to get the slope of the line:
$y = -x^3 + 2x + 1$
$y' = -3x^2 + 2$
$y' = -3 + 2$
$y' = -1$
I don't know where to go from here.
Best Answer
We first describe the normal computational approach, and then a more conceptual approach.
Computational approach: You can find the equation of the tangent line. After some work you will end up with $y=3-x$.
Substitute $3-x$ for $y$ in the equation of the curve. We end up with a cubic equation $x^3-3x+2=0$.
This equation has $x=1$ as a root. Divide $x^3-3x+2$ by $x-1$. You will get $x^2+x-2$, which factors as $(x-1)(x+2)$. That gives the $x=-2$.
Conceptual approach: Just imagine finding the equation of the tangent line, as we did above, but don't do the actual work. When we substitute for $y$ in the equation of the curve, we end up with a cubic equation of the shape $P(x)=x^3+ax+b=0$.
Since we are not doing the work, we won't find $a$ or $b$.
The tangent line at $x=1$ kisses the curve at $x=1$. Thus the equation $P(x)=0$ has $x=1$ as a double root. So the sum of the roots is $1+1+w$, where $w$ is our mystery number.
The sum of the roots of $P(x)=0$ is the negative of the coefficient of $x^2$, in this case, $0$. So $1+1+w=0$ and therefore $w=-2$.