find the equation of the line tangent to f(x)=2x^3-x^2+1 at x=2.
write in point slope form
To find the equation of the line tangent to a function at a given point, we need to determine the slope of the tangent line and the coordinates of the point of tangency.
Step 1: Find the derivative of the function f(x) with respect to x.
The derivative of f(x) will give us the instantaneous rate of change, which represents the slope of the tangent line.
f(x) = 2x^3 - x^2 + 1
To find the derivative, we differentiate each term of the function:
f'(x) = d/dx (2x^3) - d/dx (x^2) + d/dx (1)
= 6x^2 - 2x
Step 2: Find the slope of the tangent line.
Evaluate the derivative at the point x=2 to find the slope of the tangent line at that point.
f'(2) = 6(2)^2 - 2(2)
= 24 - 4
= 20
The slope of the tangent line at x=2 is 20.
Step 3: Find the coordinates of the point of tangency.
Plug x=2 into the original function f(x) to get the corresponding y-coordinate.
f(2) = 2(2)^3 - (2)^2 + 1
= 16 - 4 + 1
= 13
The point of tangency is (2, 13).
Step 4: Write the equation of the tangent line in point-slope form.
Using the slope-intercept form of a linear equation (y - y1 = m(x - x1)), we can substitute the point of tangency (x1, y1) and the slope (m) into the equation.
Using (2, 13) as the point of tangency and a slope of 20, the equation of the tangent line in point-slope form is:
y - 13 = 20(x - 2)
Simplifying, we get:
y - 13 = 20x - 40
This can be rewritten in point-slope form as:
y = 20x - 27
Thus, the equation of the tangent line to f(x) = 2x^3 - x^2 + 1 at x=2 in point-slope form is y = 20x - 27.