Showing that the tangent to the curve is also tangent to curve at another point.

**kmjt** · April 5th 2010, 11:38 AM

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.

So far this is what I did:

Subbed in 1 to the oringinal curve to find out that the point is (1,3)

Than I derived it to 4x^3 +6x^2 -6x -4 and when I subbed in 1 into the derivative I got 0.

Than I found out the equation of the line which is y=3, but i'm completely lost on what to do from there

**Archie Meade** · April 5th 2010, 01:23 PM

Originally Posted by kmjt

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.

So far this is what I did:

Subbed in 1 to the oringinal curve to find out that the point is (1,3)

Than I derived it to 4x^2 +6x^2 -6x -4 and when I subbed in 1 into the derivative I got 0.

Than I found out the equation of the line which is y=3, but i'm completely lost on what to do from there

Yes, that is correct except for a small typo.

$f'(x)=2\left(x^2+x-2\right)(2x+1)=4x^3+6x^2-6x-4$

At x=1, f'(x)=0.

x=1, y=3. The tangent at x=1 is parallel to the x-axis
(it's a local minimum).

If this tangent is tangent at another point, then there is a 2nd turning point at y=3.

Hence, we find a second solution for x, if y=3 corresponds to f'(x)=0 for more than one x.

This also means we can examine f(x) to see how many x causes f(x)=3.

$f(x)=\left(x^2+x-2\right)^2+3=3\ \Rightarrow\ \left(x^2+x-2\right)=0$

$\Rightarrow\ (x+2)(x-1)=0$

f(x)=3 for x=1 and x=-2 only.

To check the tangent, if x=-2, f'(x)=-32+16+16+8-4-4=0.

Hence, The tangent is also tangent to the curve at (-2,3).

**kmjt** · April 5th 2010, 05:54 PM

Hmm i'm kind of confused at this point. How are you going from (x^2 +x -2)^2 +3 to (x^2 +x -2)?

**helterhax** · April 6th 2010, 12:44 AM

$f(x)=(x^2 +x-2)^2 + 3 = 3 \Rightarrow (x^2+x-2)^2 = 3-3 \Rightarrow (x^2 + x-2)^2 = 0$

$(x^2 + x-2)^2 = (x+2)^2 (x-1)^2$
Also
$(x^2 + x-2) = (x+2) (x-1)$

They have the same roots, therefore it will yield the same answer, so this is why he is jumping from:
$(x^2 + x-2)^2 + 3 = 3 \Rightarrow (x^2 + x-2) = 0$

Hope this helps.

**Archie Meade** · April 6th 2010, 01:40 AM

Originally Posted by kmjt

Hmm i'm kind of confused at this point. How are you going from (x^2 +x -2)^2 +3 to (x^2 +x -2)?

Yes,

$f(x)=\left(x^2+x-2\right)^2+3=3$

means

$\left(x^2+x-2\right)^2=0\ \Rightarrow\ \left(x^2+x-2\right)\left(x^2+x-2\right)=0$

$\Rightarrow\ \left(x^2+x-2\right)=0\ \Rightarrow\ (x+2)(x-1)=0$

$x=1$

$x=-2$

**kmjt** · April 7th 2010, 02:04 PM

Why wouldn't you expand (x^2 +x -2)(x^2 +x -2)?

**Archie Meade** · April 7th 2010, 02:34 PM

Originally Posted by kmjt

Why wouldn't you expand (x^2 +x -2)(x^2 +x -2)?

Hi kmjt,

we could, but there is no need to!
it's just extra work...

$f(x)=\left(x^2+x-2\right)\left(x^2+x-2\right)+3$

If f(x)=3, then f(x)=0+3, the product of the brackets must be zero,
hence even if we multiply out the brackets, all that will still be zero
at the second point on the graph that that particular tangent touches.

Since it is the same factor multiplied by itself,
we are left to solve

$x^2+x-2=0.$

since 0(0)=0

As this is a quadratic, it gives us at most 2 values of x causing f(x)=3.

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