find an equation of the tangent line to the curve at the indicated point.
y = -(x^1/3); (8, -2)
To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and the coordinates of the given point.
Step 1: Find the derivative of the given equation to calculate the slope of the tangent line.
The derivative of the equation y = -(x^(1/3)) can be found using the power rule of differentiation. Applying the rule, we get:
dy/dx = -(1/3)x^(-2/3)
Step 2: Evaluate the derivative at the x-coordinate of the given point to find the slope.
The x-coordinate of the given point is 8. Substitute x = 8 into the derivative equation:
dy/dx = -(1/3)(8)^(-2/3)
Now we need to simplify this expression. We can rewrite (8)^(-2/3) as 1/(8)^(2/3) and then take the cube root of 8:
dy/dx = -(1/3)(1/2)
dy/dx = -1/6
So, the slope of the tangent line is -1/6.
Step 3: Use the coordinates of the given point and the slope to determine the equation of the tangent line.
We have the slope (-1/6) and the point (8, -2). We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values into the formula:
y - (-2) = (-1/6)(x - 8)
y + 2 = (-1/6)(x - 8)
y + 2 = (-1/6)x + (4/3)
To simplify:
y = (-1/6)x + (4/3) - (2)
y = (-1/6)x + (4/3) - (6/3)
y = (-1/6)x - (2/3)
Therefore, the equation of the tangent line to the curve y = -(x^(1/3)) at the point (8, -2) is y = (-1/6)x - (2/3).