Questions: Find an equation of the tangent line to the graph of (y=frac-4 xx^2+1) at the origin and at the point ((1,-2)). The tangent to the curve at the origin is (y=)

Solution Steps

To find the equation of the tangent line to the graph of \( y = \frac{-4x}{x^2 + 1} \) at a given point, we need to:

1. Compute the derivative of the function to get the slope of the tangent line.
2. Evaluate the derivative at the given points to find the slopes at those points.
3. Use the point-slope form of the equation of a line to write the equation of the tangent line.

To find the slope of the tangent line, we first compute the derivative of the function \( y = \frac{-4x}{x^2 + 1} \). The derivative is given by:

\[ y' = \frac{8x^2}{(x^2 + 1)^2} - \frac{4}{x^2 + 1} \]

Next, we evaluate the derivative at the origin \( (0, 0) \):

\[ y'(0) = -4 \]

This means the slope of the tangent line at the origin is \( -4 \).

Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (0, 0) \) and \( m = -4 \):

\[ y - 0 = -4(x - 0) \implies y = -4x \]

Now, we evaluate the derivative at the point \( (1, -2) \):

\[ y'(1) = 0 \]

This indicates that the slope of the tangent line at the point \( (1, -2) \) is \( 0 \).

Using the point-slope form again, where \( (x_1, y_1) = (1, -2) \) and \( m = 0 \):

\[ y - (-2) = 0(x - 1) \implies y = -2 \]

Final Answer