How do you find the equation of a line tangent to the function #y=-3/(x^2-25)# at (-4, 1/3)?

Question

Elijah Abram · Accepted Answer

#y=8/27x-23/27# #•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"# #y=-3/(x^2-25)=-3(x^2-25)^-1# #"differentiate using the "color(blue)"chain rule"# #•color(white)(x)d/dx(f(g(x)))=f'(g(x))xxg'(x)# #rArrdy/dx=3(x^2-25)^-2xx2x=(6x)/(x^2-25)^2# #x=-4tody/dx=-24/81=-8/27# #m=-8/27" and " (x_1,y_1)=(-4,1/3)# #rArry-1/3=-8/27(x+4)# #rArry=-8/27x-23/27larrcolor(red)" in slope-intercept form"#

Isaiah Allen · Answer

undefined To find the equation of a line tangent to a function at a given point, you can follow these steps:

1. Find the derivative of the function.
2. Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line.
3. Use the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope, to write the equation of the tangent line.

For the function y = -3/(x^2 - 25), the derivative is dy/dx = 6x/(x^2 - 25)^2.

Substituting x = -4 into the derivative, we get m = 6(-4)/((-4)^2 - 25)^2 = -24/81.

Using the point-slope form with the given point (-4, 1/3), the equation of the tangent line is y - 1/3 = (-24/81)(x + 4).