Find two distinct points on the curve y x3 x2 − x 1 that have parallel tangent lines

Given curve is "y=x^3+x^2-x+1."

"\\therefore \\frac{dy}{dx}=3x^2+2x-1"

Let the two points which have parallel tangents lines be "(x_1,y_1)" and "(x_2,y_2)."

Therefore, slope of both tangents must be equal.

"3{x_1}^2+2{x_1}-1=3{x_2}^2+2{x_2}-1\\\\\n\\Rightarrow 3({x_1}^2-{x_2}^2)+2({x_1}-{x_2})=0\\\\\n\\Rightarrow ({x_1}-{x_2})(3({x_1}+{x_2})+2)=0\\\\\n\\therefore x_1=x_2 \\ or \\ x_1+x_2=-\\frac{2}{3}"

But, we know that "x_1\\neq x_2" because than both the points will be same.

So, "\\ x_1+x_2=-\\frac{2}{3}"

Hence, we can conclude that every two points whose sum of abscissa is "-\\frac{2}{3}" will satisfy this condition.

The line $x-2y=2$ has slope $\frac12$, so you want to set the derivative of $\frac{x-1}{x+1}$ equal to $\frac12$ and solve for $x$:

$$\frac2{x^2+2x+1}=\frac12\;.$$

This would actually be just a little easier if you hadn't multiplied out the denominator; then you'd have $$\frac2{(x+1)^2}=\frac12\;,\tag{1}$$ so $(x+1)^2=4$, $x+1=\pm 2$, ...

Added: Let's recap what's going on here. You want some lines that are to be parallel to the line $x-2y=2$. That line has slope $1/2$, so your lines have to have slope $1/2$: that's how you know that they're parallel to the line $x-2y=2$. (If they had a different slope, they'd eventually cross it somewhere.) The derivative $\frac2{(x+1)^2}$ gives the slope of the tangent at any point on the curve $y=\frac{x-1}{x+1}$; we want to know where that slope is $1/2$, so we set the derivative equal to $1/2$ in $(1)$ and solve for $x$. You found that $x=-3$ or $x=1$, so there are two points on the curve where the tangent is parallel to the line $x-2y=2$; one is $(-3,2)$, and the other is $(1,0)$. Now you just find the tangent lines to $y=\frac{x-1}{x+1}$ at those two points.

For the first, for instance, you have $y-2=\frac12\big(x-(-3)\big)=\frac12(x+3)=\frac12x+\frac32$, so $y=\frac12x+\frac72$; I'll leave the other to you.

I glanced at Jim's answer, which looks like a nice, standard calculus treatment. But I couldn't help but feel sad for all the middle schoolers out there in Socratic land who want to find tangents of algebraic curves but are still years away from calculus.

Fortunately they can do these problems using only Algebra I.

#x^2+xy+y^2=7#

This might be a bit complicated for a first example, but let's go with it. We write our curve as #f(x,y)=0# where

#f(x,y) = x^2+xy+y^2-7#

Let's take #(r,s)# as a point on #f#. We want to investigate #f# near #(r,s)# so we write

#f(x,y)=f(r + (x-r), s + (y-s))#

# = (r + (x-r)) ^2+(r + (x-r))( s + (y-s))+( s + (y-s))^2-7#

We expand, but we don't expand the difference terms #x-r# and #y-s#. We want to keep those intact so we can experiment with eliminating some later.

#f(x,y) = r^2 + 2r(x-r) + (x-r) ^2+ (rs + s(x-r) + r(y-s) + (x-r)(y-s))+s^2 + 2s(y-s) + (y-s)^2-7#

#= ( r^2 + rs + s^2 - 7) + (2r+s)(x-r) + (2s+r)(y-s) + (x-r) ^2+ (y-s)^2+ (x-r)(y-s)#

#= f(r,s) + (2r+s)(x-r) + (2s+r)(y-s) + (x-r) ^2+ (y-s)^2+ (x-r)(y-s)#

We said #(r,s)# is on #f# so #f(r,s)=0#.

#f(x,y) = (2r+s)(x-r) + (2s+r)(y-s) + (x-r) ^2+ (y-s)^2+ (x-r)(y-s)#

We sorted the terms by degree, and we can experiment with approximations to #f# near #(r,s)# by dropping the higher degrees. The idea is when #(x,y)# is near #(r,s)# then #x-r# and #y-s# are small, and their squares and product are smaller still.

Let's just generate some approximations to #f#. Since #(r,s)# is on the curve, the constant approximation, dropping all the difference terms, is

#f_0(x,y) = 0#

That's not particularly exciting, but it correctly tells us points near #(r,s)# will give a value near zero for #f#.

Let's get more interesting and keep the linear terms.

#f_1(x,y) = (2r+s)(x-r) + (2s+r)(y-s) #

When we set this to zero, we get the best linear approximation to #f# near #(r,s),# which is the tangent line to #f# at #(r,s).# Now we're getting somewhere.

# 0 = (2r+s)(x-r) + (2s+r)(y-s) #

We can consider other approximations as well:

#f_2(x,y) = (2r+s)(x-r) + (2s+r)(y-s) + (x-r) ^2 #

#f_3(x,y) = (2r+s)(x-r) + (2s+r)(y-s) + (x-r) ^2 + (x-r)(y-s)#

These are higher order tangents, ones that college math students hardly ever get to. We've already gone beyond college calculus.

There are more approximations, but I'm being warned this is getting long. Now that we learned how to do calculus using only Algebra I, let's do the problem.

We want to find the points where the tangent line is parallel to the #x# axis and #y# axis.

We found our tangent line at #(r,s)# is

# 0 = (2r+s)(x-r) + (2s+r)(y-s) #

Parallel to the #x# axis means an equation #y = text{constant}#. So the coefficient on #x# must be zero:

#2r + s = 0#

#s = -2r#

#(r,s)# is on the curve so #f(r,s)=0#:

# r^2 + rs + s^2 - 7 = 0 #

#r^2 + r(-2r) + (-2r)^2 - 7 = 0#

#r =pm sqrt{7/3}#

Since #s=-2r# the points are

#(-sqrt{7/3}, 2sqrt{7/3}) and (sqrt{7/3}, -2sqrt{7/3}) #