Finding equations of the tangent lines

Question

Find equations of the tangent lines to the curve that are parallel to the line x − 2y = 4.

 (Enter your answers as a comma-separated list of equations.)

Hi, can someone please explain the steps to solving this problem including finding tangent lines to the curve?

Answer 1

If you place the given line in slope intercept form: y = (1/2)x - 2, it becomes evident that the slope of that line is 1/2. That means you want the slope of the tangent lines to be 1/2 (parallel lines have the same slope).

Since y' is a "formula" for slopes of tangent lines to the graph of the original function, set y' equal to 1/2 and solve for x. This will give the x-coordinates of the points on the graph that have tangent lines with a slope of 1/2.

2/(x+1)² = 1/2

(x+1)² = 4

x + 1 = ±2

x₁= -3 x₂= 1

Substituting those values into the original function gives the y-coordinates of the points of tangency.

f(-3) = [(-3-1)]/[-3+1] = -4/-2 = 2

So, (-3,2) is on the graph of the original function and the tangent line at that point has a slope of 1/2.

f(1) = 0

So (1,0) is on the graph and has a tangent line with a slope of 1/2.

In order to write the equations you have a point and slope.

When x = -3, y = 2

y - 2 = (1/2)(x +3)

y = (1/2)x + 7/2

When x = 1, y = 0

y = (1/2)(x-1)

y = (1/2)x - 1/2

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Here is a graph that should help visualize. The light green dashed line is the graph of the derivative.

desmos.com/calculator/1iflyyops3

Answer 2

Here's the Desmos animation to go along with it: https://www.desmos.com/calculator/irdrdhqy3p

Answer 3

To find a tangent line to a curve, we need 2 things: the slope (expressed as the function’s derivative), and a point on the curve given by:

y = f(x) = (x - 1) / (x + 1)

With rational expressions of polynomials such as this one, to find the derivative, we always have (I) the Quotient rule for differentiation, but (II) in simpler cases like this one, we are reminded of the much simpler relationship d/dx ( 1/x ) = -1 / x²

In this case: y = (x - 1) / (x + 1)

Quotient Rule: y’ = [ (x + 1)(1) - (x - 1)(1) ] / (x + 1)² = 2 / (x + 1)² # Careful with all those signs !

Simplification: y = (x - 1) / (x + 1) = (x + 1 - 2) / (x + 1) = (x + 1)/(x + 1) - 2/(x + 1)= 1 - 2 / (x + 1)

So we can think of our curve as: y = 1 - 2 / (x + 1). Besides simplifying the derivative, this helps with visualization, as we can now think of this curve as: y = 1/x, vertical axis shifted to x = -1, exaggerated/stretched by factor of 2, flipped upside down by -1, then all shifted up by 1. Do you see this ? Now, checkout Desmos… entering each step I just outlined… y = 1/x => 1/(x+1) => -2/(x + 1) => 1 - 2/(x + 1) Do you see the progression ?

==> Remembering: d/dx (1/x) = -1/x², y’ = d/dx [ -2 / (x + 1) ] = (-1)(-2) / (x + 1)² = 2 / (x + 1)²

# There is no concern over the Chain Rule here, as d/dx (x) = 1… otherwise, what might happen with y = 1 - 2/(ax + b) ?

In our case, we are not given (I)… a point for our tangent line, but instead we are given another (parallel) line, from which we can (a) extract the desired slope (we don’t care about the rest), then (b) we use that slope to see where on our curve do candidate values of ‘x’ produce that known slope… there may be more than 1… we’ll check for validity.

(a) We are given a line: x − 2y = 4 ==> y = (x - 4) / 2 , so our coefficient of x will be m = 1/2 for the desired slope. This is all we need from here.

(b) From dy/dx = 2 / (x + 1)² , we want dy/dx = slope 1/2 ==> 2/(x + 1)² = 1/2 ==> (x + 1)² = 4

==> x + 1 = ± 2 ==> x = -1 ± 2 ==> x = 1 or -3 , and from y = f(x) we now have points (1, 0), (-3, -4/-2) = (-3, 2)

Given the graph (or mental image) above, of our curve, there are no problems in the neighborhood of these 2 points.

At (1, 0) with slope 1/2 ==> (y - 0) / (x - 1) = m = 1/2 ==> y = (1/2)(x - 1) ==> y = x/2 - 1/2

At (-3, 2) with slope 1/2 ==> (y -2) / (x - (-3)) = m = 1/2 ==> y = (1/2)(x + 3) + 2 = x/2 + (2 + 3/2) ==> y = x/2 + 7/2

[Attention to Detail: The question asks for a “comma-separated list of equations". OK, fine, we even serve fries with that !

Answer: y = x/2 - 1/2 , y = x/2 + 7/2

Answer 4

y= (x-1)/(x+1)

derivative =

y'=2/(x+1)^2 = slope of the tangent line =1/2 as all parallel lines have the same slope = coefficient of the x term when a linear equation is solved for y

2/(x+1)^2 = 1/2

cross multiply

4 = (x+1)^2

solve for x, take square roots of both sides to get

x +1 = +/-2

x =-1 +/-2= -3 or +1

y= (x-1)/(x+1) = -4/-2 = 2 or 0

x,y = (1,0) or (-3,2)

two tangent lines

one is y = x/2-1/2

2nd is y-2 = (1/2)(x+3)

or y = x/2 +7/2