The curve y = x/(1 + x^2)is called a serpentine. Find an equation of the tangent line to this curve at the point (4, 0.24).

Round the slope and y-intercept to two decimal places.)

first cross-multiply

y + y x^2 = x
dy/dx + y(2x) + 2x dy/dx = 1
dy/dx(1 + 2x) = 1 - 2xy
dy/dx = (1-2xy)/(1+2x)
at (4, 1/4)
dy/dx = (1 - 2(4)(1/4))/(1 + 2(4))
= -1/9

equation of tangent:
y = (-1/9)x + b
at (4,1/4)
1/4 = (-1/9)(4) + b
b = 5/9

equation is y = (-1/9)x + 5/9

slope = -1/9
y-intercept is 5/9

Why would you round off and create inaccuracy, while the above gives you exact values.

BTW, just noticed this same question 3 years ago here
http://www.jiskha.com/display.cgi?id=1320285563

(I answered it by taking the derivative directly, and they changed the y value of the given point)

Well, the tangent line to the serpentine curve at (4, 0.24) is like a snake slithering along at that point. Let's find its equation!

First, we need to find the slope of the tangent line. To do that, let's take the derivative of the curve equation. Now, don't worry, the only "d" you need to know here is that "d" stands for "derivative"!

Taking the derivative of y = x/(1 + x^2) requires a bit of patience, but remember, patience is a virtue, just like any skill required for tangents to serpentine curves.

Using the quotient rule, it turns out that the derivative of this equation is (1 - x^2)/(1 + x^2)^2. Phew!

Now, plug in the x-coordinate of the point we have, which is 4, into the derivative equation. After some calculations, we find that the slope of the tangent line at (4, 0.24) is approximately -0.08.

Next up, we need to determine the y-intercept of the tangent line. To do that, we can use the point-slope form of a line y - y₁ = m(x - x₁), where (x₁, y₁) is (4, 0.24) and m is the slope we found (-0.08).

Substituting the values into the equation, we get y - 0.24 = -0.08(x - 4). Now, let's simplify!

Distributing the -0.08, we get y - 0.24 = -0.08x + 0.32. Almost there!

Finally, let's isolate y by adding 0.24 to both sides, resulting in y = -0.08x + 0.56. Ta-da! The equation of the tangent line to the serpentine curve at (4, 0.24) is y = -0.08x + 0.56.

So, the slope is approximately -0.08, and the y-intercept is 0.56 (rounded to two decimal places). But remember, humor always rounds up, so it's like the slope is saying, "0.08, I'm not negative, just a little down on the curve!" And the y-intercept is like, "0.56, I'm up here being supportive, ready to boost your confidence!"

To find the equation of the tangent line to the curve y = x/(1 + x^2) at the point (4, 0.24), we'll need to follow these steps:

Step 1: Find the derivative of the curve y = x/(1 + x^2).
Step 2: Plug in the x-coordinate of the given point to find the slope of the tangent line.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.

Let's proceed with these steps one by one.

Step 1: Find the derivative of the curve y = x/(1 + x^2).

To find the derivative, we can use the quotient rule. Let u = x and v = 1 + x^2.

Applying the quotient rule: (u/v)' = (vu' - uv') / v^2

u' = 1 (derivative of x)
v' = 2x (derivative of 1 + x^2)

Now, let's substitute these values into the quotient rule:

(y)' = ((1 + x^2)(1) - (x)(2x)) / (1 + x^2)^2
= (1 + x^2 - 2x^2) / (1 + x^2)^2
= (1 - x^2) / (1 + x^2)^2

So the derivative of y = x/(1 + x^2) is (y)' = (1 - x^2) / (1 + x^2)^2.

Step 2: Plug in the x-coordinate of the given point to find the slope of the tangent line.

To find the slope of the tangent line, substitute x = 4 into the derivative:

slope = (1 - 4^2) / (1 + 4^2)^2
= (1 - 16) / (1 + 16)^2
= -15 / 17^2
≈ -0.0518 (rounded to two decimal places)

Step 3: Use the point-slope form of a line to find the equation of the tangent line.

We have the slope (-0.0518) and the point (4, 0.24).

Using the point-slope form of a line: y - y1 = m(x - x1)

Plugging in the values: y - 0.24 = -0.0518(x - 4)

Now, let's simplify and rewrite the equation in the standard form:

y - 0.24 = -0.0518x + 0.2072
y = -0.0518x + 0.4472

So, the equation of the tangent line to the curve y = x/(1 + x^2) at the point (4, 0.24) is y = -0.0518x + 0.4472.

To find the equation of the tangent line to the curve at the point (4, 0.24), we need to determine the slope of the tangent line at that point.

Step 1: Determine the derivative of the curve.
The derivative of a function represents the slope of the tangent line at any given point on the curve. So, we need to find the derivative of the curve y = x/(1 + x^2).

To find the derivative, we can use the quotient rule, which states that the derivative of a quotient of two functions is given by (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2.

Let's apply the quotient rule to find the derivative of y = x/(1 + x^2):
f(x) = x
g(x) = 1 + x^2

f'(x) = 1 (derivative of x is 1)
g'(x) = 2x (derivative of 1 + x^2 is 2x)

Using the quotient rule, the derivative of y = x/(1 + x^2) is:
y' = ((1)(1 + x^2) - (x)(2x)) / (1 + x^2)^2
y' = (1 + x^2 - 2x^2) / (1 + x^2)^2
y' = (1 - x^2) / (1 + x^2)^2

Step 2: Evaluate the derivative at the point (4, 0.24).
To find the slope of the tangent line at the point (4, 0.24), we substitute x = 4 into the derivative expression:
y' = (1 - (4)^2) / (1 + (4)^2)^2
y' = (1 - 16) / (1 + 16)^2
y' = (-15) / (17)^2
y' = -15 / 289

Therefore, the slope of the tangent line at the point (4, 0.24) is -15 / 289 (rounded to two decimal places).

Step 3: Use the point-slope form of the line to find the equation of the tangent line.
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Using the point (4, 0.24) and the slope -15 / 289, we can write the equation of the tangent line as:
y - 0.24 = (-15 / 289)(x - 4)

Simplifying:
y - 0.24 = (-15/289)x + (60/289)
y = (-15/289)x + (60/289) + 0.24
y = (-15/289)x + (60/289) + (69/289)
y = (-15/289)x + (129/289)

Therefore, the equation of the tangent line to the curve at the point (4, 0.24) is y = (-15/289)x + (129/289).