Let f be the function defined by f(x) = 2{x^3} - 4x + 1. Which of the following is an equation of the line tangent to the graph of f at the point x = - 1?
I don't see any choices.
y = 2x^3 - 4x + 1
dy/dx = 6x^2 - 4
at x = -1, dy/dx = 6-4 = 2
y = -2 + 4+1 = 3
so we have slope = 2, point is (-1,3)
equation of tangent:
y - 3 = 2(x + 1)
dress it up to whatever form you need.
To find the equation of the tangent line to the graph of f at the point x = -1, we need to find the slope of the tangent line and the coordinates of the point of tangency.
The slope of the tangent line can be found by taking the derivative of f(x). Let's first find the derivative of f(x):
f(x) = 2x^3 - 4x + 1
To find the derivative, we apply the power rule for differentiation:
f'(x) = d/dx [2x^3] - d/dx [4x] + d/dx [1]
= 6x^2 - 4
Now, to find the slope of the tangent line at x = -1, we substitute x = -1 into the derivative:
f'(-1) = 6(-1)^2 - 4
= 6 - 4
= 2
So, the slope of the tangent line is 2.
To find the coordinates of the point of tangency, we substitute x = -1 into the original function f(x):
f(-1) = 2(-1)^3 - 4(-1) + 1
= -2 + 4 + 1
= 3
Therefore, the coordinates of the point of tangency are (-1, 3).
Now, we have the slope of the tangent line (2) and a point on the line (-1, 3). We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is a point on the line.
Plugging in the values, we have:
y - 3 = 2(x - (-1))
y - 3 = 2(x + 1)
y - 3 = 2x + 2
y = 2x + 5
Therefore, the equation of the line tangent to the graph of f at the point x = -1 is y = 2x + 5.