# Calculus: The tangent line intersects a curve at two points. Find the other point.

## 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.

First of all, it is $f(1)=2$, so the tangent line is passing through the point $(1,2)$ and it's slope is $k=f'(1)=-3\dot{}1+2=-1$. So the tangent line is given by the equation $t(x)=kx+c=-x+c$ for some $c\in{}R$. We can easily determine that $c$ from the fact, that the tangent line is pasing throught $(1,2)$. We get $$t(x)=-x+3$$ Now we are looking for all $x\in{}R$ such $t(x)=f(x)$. This equation is equivalent to $$0=t(x)-f(x)=x^3-3x+2=(x-1)(x^2+x-2)=(x-1)^2(x+2)$$ So $x=1\vee{}x=-2$ and so the tanget line intersects the curve at $(1,2)$ (which we already know, of course) and at $(-2,5)$.

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 $w=-2$.

Your statement that "This was never taught in class" might astonish your instructor. But even if not, it is very unreasonable to expect to be required to do only that which someone has shown you how to do. And this is so close to the beaten path that it's not a good example of something you might not have been shown how to do.

When $x=1$, then $y=2$ so you've got a line passing through $(1,2)$ with slope $-1$. In earlier courses you learned how to write an equation for that line.

Now you need the point of intersection of the graphs of two equations, $y=−x^3+2x+1$ and one other. You've probably seen problems like that before, and if not, apply some common sense, and if that doesn't work, then tell us with specificity at what point you ran into difficulty.

• Actually, as my teacher was handing out the worksheet, she told us that people in her other classes were confused by this question as well. She gave us the worksheet at the end of class, so I didn't really look at it until I sat down to actually do the assignment. – Amanda Sep 8 '13 at 22:20
• @Amanda : This is an example of the fact that large numbers of students who don't sufficiently know the prerequisite material are pressured into taking calculus. It's a racket: it brings in tuition money. One important thing you should have learned in previous courses is that math problems are often like puzzles: You haven't been told how to solve them, but you've been given all the information you need to figure it out if you think about it. But students in calculus courses often feel they're entitled to be told in advance how to do problems. That's wrong. – Michael Hardy Sep 9 '13 at 17:49

If we let $f(x)=-x^3+2x+1$, then we know the derivative at $x=1$, namely $f'(1)=-1$. Now we know the slope of the tangent line $t(x)$. In particular, this implies $t(x)=-x+c$ for some constant $c \in \mathbb{R}$.

We can also compute a point on the line, namely $(1,f(1))$. This gives the value of $t(1)=f(1)$, which can be used to solve for $c$.

Then find the other point where $f(x)$ and $t(x)$ intersect, i.e., where $f(x)-t(x)=0$.

Hint: a line satisfying the "point-slope form" equation $(y-y_0)=m(x-x_0)$ goes through point $(x_0,y_0)$ and has slope $m$. In the case of our tangent line, $$x_0=1 \\ y_0=f(x_0)=f(1)$$ and $$m=f'(1)=-1$$ (as you correctly computed).

Now that you have the two curves, set them equal to solve for their intersections.

You can find the equation of the line tangent when $x=1$. You already have its slope, wich is $-1$, then if $f(x)$ is the such function

$$f(x)=mx+b=-x+b,$$

to find $b$, you need see that this line must pass through point $(1, -(1)^3+2(1)+1)=(1,2)$ and then you can find $b$. Therefore you can solve the equation

$$-x+b=-x^3+2x+1$$

to find all values $x$ of intersection, including $x=1$.

x=1

y=-x3+2x+1

y=2

m=dy/dx=-3x2+2

x=1

m=-1

y-2=-1(x-1)

y=3-x

3-x=-x3+2x+1

0=-x3+3x-2

let x=1

0=0

:: x-1 is a factor

(x-1)(-x2-x+2)=-x3+3x-2

-x2-x+2=0

x=1 , x=-2

0=(x-1)(x-1)(x+2)

x=1(N.A) , x=-2 , x=1(N.A)

y=5