I need help. I have no clue about how to solve this problem. Can someone go through it.
Find the sope of the tangent line to the graph of the equation
y=x^2(x-3) at x=-1
Thanks for the help.
the slope of the tangent line at a given point equals the value of the first derivative at that point, so ...
y' = 3x^2 - 6x
when x = -1
y' = 3 - (-6) = 9
(in your next lesson you will probably find the equation of that tangent)
To find the slope of the tangent line to the graph of the equation, we can follow the steps below:
Step 1: Start with the equation given:
y = x^2(x - 3)
Step 2: Differentiate the equation with respect to x to find the derivative. The derivative represents the rate of change of y with respect to x and gives us the slope of the tangent line at any given point.
dy/dx = 2x(x - 3) + x^2(1)
Step 3: Simplify the derivative equation:
dy/dx = 2x^2 - 6x + x^2
= 3x^2 - 6x
Step 4: Evaluate the derivative equation at the specified value of x, which in this case is x = -1. Substituting x = -1 into the derivative equation, we get:
dy/dx = 3(-1)^2 - 6(-1)
= 3 - (-6)
= 3 + 6
= 9
Step 5: The result obtained in Step 4 represents the slope of the tangent line to the graph of the equation at x = -1. Therefore, the slope of the tangent line is 9.
So, the slope of the tangent line to the graph of the equation y = x^2(x - 3) at x = -1 is 9.