Suppose f(x)=5/(x-2). f'(5)=-5/9. Use this to find the equation of the tangent line to the curve y=5/x-2 at the point (5,(5/3)). write your answer in the form y=mx+b where m is the slope and b is the y-intercept. The equation of the tangent line is.....
equation of tangent:
y - 5/3 = (-5/9)(x - 5)
y = (-5/9)x + 25/9 + 5/3
y = (-5/9)x + 40/9
check my arithmetic
To find the equation of the tangent line to the curve y = 5/(x - 2) at the point (5, 5/3), we need to determine both the slope and the y-intercept of the line.
First, we can find the slope of the tangent line using the derivative. Given that f'(x) = -5/9, we substitute x = 5 into the derivative:
f'(5) = -5/9
This tells us that the slope of the tangent line at the point (5, 5/3) is -5/9.
Next, we can determine the equation of the tangent line using the point-slope form of a linear equation: y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, and m is the slope.
Substituting the given point (5, 5/3) and the slope -5/9 into the equation gives:
y - 5/3 = (-5/9)(x - 5)
To put it in the form y = mx + b, we can simplify:
y - 5/3 = (-5/9)(x - 5)
y - 5/3 = (-5/9)x + 25/9
y = (-5/9)x + 25/9 + 5/3
y = (-5/9)x + 25/9 + 15/9
y = (-5/9)x + 40/9
Therefore, the equation of the tangent line is y = (-5/9)x + 40/9, where the slope (m) is -5/9, and the y-intercept (b) is 40/9.