find the equationss of both tangenr lines to the curve defined by y=(1/x^2), x cannot equal 0 from the external point P(0,1).

I tried but i got one equation. Can you please verify if it is: y=-0.38x-1.

Thanks

To find the equations of the tangent lines to the curve y = 1/x^2 from the external point P(0,1), we can use the following steps:

Step 1: Find the derivative of the function y = 1/x^2 using the power rule for differentiation.
dy/dx = -2/x^3

Step 2: Substitute the x-coordinate of the external point P into the derivative to find the slope of the tangent line passing through P.
m = dy/dx|P = -2/(0^3) = undefined

Since the slope is undefined, it means that the tangent line is vertical. The equation of a vertical line passing through P(0,1) is simply x = 0.

Step 3: To find the other tangent line, we need to find the slope using a different approach. Since the slope of the tangent line equals the derivative of y with respect to x, we can set dy/dx = m and solve for x.

m = -2/x^3
m*x^3 = -2
x^3 = -2/m
x = (-(2/m))^(1/3)

Step 4: Substitute the value of x into the equation y = 1/x^2 to find the corresponding y-coordinate.
y = 1/[(-(2/m))^(1/3)]^2
= 1/[(-(2^(1/3))/m)^(2)]
= m^2/(2^(2/3))

The coordinates of the point where the second tangent line touches the curve are (-(2/m))^(1/3), m^2/(2^(2/3)).

Step 5: Now, we have the slope (m) and a point (-(2/m))^(1/3), m^2/(2^(2/3)). We can use the point-slope form of the linear equation to find the equation of the second tangent line.
y - m^2/(2^(2/3)) = m(x - (-(2/m))^(1/3))

Simplifying this equation will give you the equation of the second tangent line.

Please note that the equation you provided, y = -0.38x - 1, is not the correct equation for either of the tangent lines.