Finding points where a tangent lines is parallel to another line

**hazecraze** · October 15th 2009, 06:32 AM

#8. find each point at which the tangent line to the curve $y1=2x +(4/x) +1$ is parallel to the line $y2+2x=6$ .

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).

**stapel** · October 15th 2009, 06:42 AM

Solve the second equation for "y=". Then read the slope value off the equation. (It'll be the value multiplied on the "x".)

Differentiate the first equation with respect to x. Set equal to the given slope value, and solve for the corresponding x-value(s).

Plug these x-values into the first equation, and solve for the corresponding y-value(s), and thus the point(s) in question.

(I get the same answer as you've listed.)

**hazecraze** · October 15th 2009, 08:04 AM

At first I confused my self by thinking it was $y1=2x +(4/x +1)$ and got (1,3), but I see that it was $y1=2x +(4/x) +1$ . Also, after differentiating and setting equal to -2 and then simplifying, I got $-(x^3+2x^2+x+1)$ . With synthetic division and a $(x-1)$ root, I got -(x-1)=0, which was the $x$ value of (1,7), but for the other part: $(x^2+x)$ , I factored out an $x$ to get $x(x+1)$ , so wouldn't there also be a point 0, in addition to the -1, thus giving (0,1) and then the (1,7) and (-1,-5)?

**ramiee2010** · October 15th 2009, 08:40 AM

Originally Posted by hazecraze

#8. find each point at which the tangent line to the curve $y1=2x +(4/x) +1$ is parallel to the line $y2+2x=6$ .

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).

$slop\ of\ line \ y_{2}+2x=6 is m_{2}=-2\quad (\because y=mx+c)$
$slop\ of \ curve \ y_{1}= 2x +\frac{4}{x} +1 \ is$
$m_{1}=\frac{dy}{dx}=\quad2- \frac{4}{x^2}$
$since\ the\ tangent\ line\ y_{1}\ to\ the\ curve\ is\ parallel\ to\ the\ line\ y_{2}$
$m_{1}=m_{2} \quad \ so \ that\ \quad \quad2- \frac{4}{x^2} = -2$
$or \ x= \pm 1$
on substituting in equation of curve we get
$for \ x=1 \quad y=7 \ and\ for \ x=-1 \quad y=-5$
at (1, 7) and (-1, -5) ,the tangent line to the curve $\ y_{1}= 2x +\frac{4}{x} +1$ is parallel to the line $y_{2}+2x=6$

**hazecraze** · October 15th 2009, 09:31 AM

Oh, well that was much more simplistic than what I did.

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