which of the following is an equation of the line tangent to the graph of y = (x^3)-x at the point where x =2:
a) y - 6 = 4(x-2)
b) y - 6 = 5(x-2)
c) y - 6 = 6(x-2)
d) y - 6 = 11(x-2)
e) y - 6 = 12(x-2)
thanks!
D.
Plugging 2 into original fxn gives you y-coordinate. (2,6)
Taking the derivative of y=(x^3)-x gives you f'(x)=3x^2-1. Plug in x=2 to find slope. use point slope form to get equation! :)
Well, let's put on our tangent caps and see which equation fits the line at x = 2! To find the equation of the tangent line, we need to find the slope of the line at that point. So, let's take the derivative of y = (x^3) - x.
The derivative of y = (x^3) - x is 3(x^2) - 1. Now, let's substitute x = 2 into the derivative to find the slope at that point.
When we substitute x = 2: 3(2^2) - 1 = 11.
So, the slope of the tangent line at x = 2 is 11. Now let's find the equation of the line using the point-slope form!
The point (2, f(2)) is on the line, so it's (2, (2^3) - 2) = (2, 6).
Now we can use the point-slope form, with slope 11 and point (2, 6):
y - 6 = 11(x - 2).
If you simplify the equation, it becomes y - 6 = 11x - 22.
So, the correct answer is d) y - 6 = 11(x - 2).
Hope this helps and brings a smile to your face!
To find the equation of the line tangent to the graph of y = (x^3) - x at the point where x = 2, we need to calculate the derivative of the function at that point and use it to form the equation of the tangent line.
Step 1: Find the derivative of y = (x^3) - x.
To find the derivative, we take the derivative of each term separately. The derivative of x^3 is 3x^2, and the derivative of -x is -1. Therefore, the derivative of y = (x^3) - x is dy/dx = 3x^2 - 1.
Step 2: Substitute x = 2 into the derivative equation.
Substitute x = 2 into dy/dx = 3x^2 - 1 to find the slope of the tangent line at x = 2:
dy/dx = 3(2^2) - 1 = 3(4) - 1 = 12 - 1 = 11.
Step 3: Construct the equation of the tangent line.
Now that we have the slope of the tangent line (11) and the point (2, y) at which it intersects the graph, we can use the point-slope form of the equation of a line:
y - y1 = m(x - x1), where m is the slope and (x1, y1) are the coordinates of a point on the line.
Substitute the values we have into this equation:
y - 6 = 11(x - 2).
Therefore, the correct equation of the line tangent to the graph of y = (x^3) - x at the point where x = 2 is:
d) y - 6 = 11(x - 2).