Find an equation of the tangent line to the curve of f (x) = x^3 − 8, at the point where the curve crosses the x-axis.
f ' (x) = 3x^2
when the curve crossess the x-axis, y = 0
x^3 - 8 = -
x^3 = 8
x = 2 , so we have the point (2,0)
when x = 2, slope = 3(4) = 12
y - 0 = 12(x-2)
y = 12x - 24
To find the equation of the tangent line to the curve of f(x) = x^3 - 8 at the point where the curve crosses the x-axis, we need to determine the slope of the tangent line at that point.
Since the curve crosses the x-axis at this point, it means f(x) = 0. Therefore, we need to find the x-coordinate where f(x) = 0.
Setting f(x) = x^3 - 8 to zero:
0 = x^3 - 8
Adding 8 to both sides:
8 = x^3
Taking the cube root of both sides:
∛8 = ∛x^3
Simplifying:
2 = x
Now that we have the x-coordinate of the point where the curve crosses the x-axis, we can find the slope of the tangent line at that point.
The slope of the tangent line can be found by taking the derivative of f(x) with respect to x and evaluating it at x = 2.
f'(x) = 3x^2
Evaluating at x = 2:
f'(2) = 3(2)^2 = 12
So, the slope of the tangent line at x = 2 is 12.
Now that we have the point (2, 0) and the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line.
Point-slope form: y - y1 = m(x - x1)
Plugging in the values:
y - 0 = 12(x - 2)
Simplifying:
y = 12x - 24
Therefore, the equation of the tangent line to the curve of f(x) = x^3 - 8 at the point where the curve crosses the x-axis is y = 12x - 24.
To find the equation of the tangent line to the curve of f(x) = x^3 - 8 at the point where the curve crosses the x-axis, we need to find the point where the curve intersects the x-axis and the slope of the tangent line at that point.
Step 1: Find the x-intercept of the curve:
Setting f(x) = 0 gives us:
x^3 - 8 = 0
Step 2: Solve for x:
We can rewrite the equation as:
x^3 = 8
Taking the cube root of both sides, we get:
x = ∛8
Simplifying, we find:
x = 2
So the curve intersects the x-axis at x = 2.
Step 3: Find the derivative of the function f(x):
To find the slope of the tangent line, we need to find the derivative of the function f(x) = x^3 - 8.
Taking the derivative, we get:
f'(x) = 3x^2
Step 4: Evaluate the derivative at the x-coordinate where the curve intersects the x-axis:
Substituting x = 2 into f'(x), we have:
f'(2) = 3(2)^2 = 12
So the slope of the tangent line at the point where the curve crosses the x-axis is 12.
Step 5: Write the equation of the tangent line:
Using the point-slope form, we have the equation of the tangent line:
y - f(2) = f'(2)(x - 2)
Substituting x = 2 and f(2) = 0, we get:
y - 0 = 12(x - 2)
Simplifying, we find:
y = 12x - 24
Therefore, the equation of the tangent line to the curve of f(x) = x^3 - 8 at the point where the curve crosses the x-axis is y = 12x - 24.