Find an equation of the tangent line to the curve f(x)=-1/x+2 at the points (2, -1/4).
is it -1/(x+2) or (-1/x) + 2
In either case, take the derivative, put in (x=2), and that is the slope.
Y=mx+b
now at the point in question, you have, x, m, y solve for b.
f'(x) = -1/x^2
which at (2, -1/4) is -1/4
so y + 1/4 = -1/4(x-2)
times 4
4y + 1 = -(x-2)
x+4y = 1
Bobpursley, It is -(1)/(x+2)
To find the equation of the tangent line to the curve at a given point, we need to determine the slope of the tangent line and then use the point-slope form of a line.
Step 1: Find the derivative of the function f(x)=-1/x+2.
The derivative of f(x)=-1/x+2 can be found using the power rule and the chain rule. Applying the power rule, the derivative of -1/x is 1/x^2. The derivative of 2 is 0 since it's a constant.
Therefore, the derivative of f(x)=-1/x+2 is f'(x)=1/x^2.
Step 2: Evaluate the derivative at x=2.
To find the slope of the tangent line at a given point, we substitute x=2 into the derivative equation f'(x)=1/x^2.
So, f'(2)=1/2^2=1/4.
Step 3: Use the point-slope form of a 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 of the line.
We are given the point (2, -1/4) and the slope m=1/4.
Substituting this information into the point-slope form, we have: y - (-1/4) = (1/4)(x - 2)
Simplifying this equation yields: y + 1/4 = 1/4(x - 2)
Thus, the equation of the tangent line to the curve f(x) = -1/x + 2 at the point (2, -1/4) is y + 1/4 = 1/4(x - 2).