# Homework Help: More than one tangent

1. Jan 31, 2010

### fghtffyrdmns

1. The problem statement, all variables and given/known data
Show that the tangent to the curve $$y=(x^2+x-2)^2+3$$ at the point where x =1 is also tangent to the curve at another point.

2. Relevant equations

$$y=(x^2+x-2)^2+3$$

3. The attempt at a solution

$$y'=2(x^2+x-2)(2x+1)$$
$$y'(1)=0$$
$$0=2(2x+1)(x+2)(x-1)$$

x= -1/2, -2, 1

Would this be correct?

Last edited: Jan 31, 2010
2. Jan 31, 2010

### MarcMTL

Yes.

For $$f'(x) = 0$$, I also get x= -2 or x= -1/2 or x = 1.

3. Jan 31, 2010

### fghtffyrdmns

Something does not seem right to me though. I found y'(1) = 0. Therefor the tangent line at x=1 is a horizontal line. I don't understand why x=-1/2 is an answer. I got -2 as the other point.

4. Jan 31, 2010

### MarcMTL

I graphed out the function in Maple to make sure.

You can clearly see where the derivative would be = 0, at the 3 points quoted above.
Hope this helps out.

5. Jan 31, 2010

### fghtffyrdmns

Yes, this is how why I am confused. The tangent line at x=1 is y=3. That is also the same at x=-2. But why -1/2?

I'm confusing myself :/.

6. Jan 31, 2010

### Staff: Mentor

Because the tangent at x = -1/2 is horizontal, thus y'(-1/2) = y'(-2) = y'(1) = 0. The tangent line at x = -1/2 has a different equation than those at x = -2 and x = -1.

Setting the derivative equal to zero tells you where the function has horizontal tangents, but it doesn't tell you the equations of the tangent lines.

BTW, the graph that MarcMTL provided is very rough. At the turning points, the graph of the function is much smoother.

7. Jan 31, 2010

### fghtffyrdmns

So my answers the first time are correct?
I think I misinterpreted the question as to where the tangent at 0 crosses another point.

8. Jan 31, 2010

### Staff: Mentor

The values are correct, but I don't think you answered the question you were supposed to answer; namely, show that the tangent at x = 1 is also tangent to the curve at another point. You should have said something about the tangent line at x = 1 and the tangent line at x = -2 both have the same equation: y = 3. The problem isn't at all concerned with the tangent at x = -1/2.

9. Jan 31, 2010

### LCKurtz

Did you really plot that graph in Maple? It should look much nicer:

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