Find the slope of the tangent line to the graph of the equation:
y = (2x^2+3)(x-1)
at x=2.
a = 11
b = 08
c = 19
d = 16
I would expand it, then take the derivative
then subbing in x=2 can be done mentally
I did that and got 19. Is this right?
To find the slope of the tangent line to the graph of the equation at a specific point, we need to take the derivative of the function and evaluate it at that point.
Given the equation: y = (2x^2 + 3)(x - 1), we will first find its derivative using the product rule of differentiation.
Let's start by expanding the equation:
y = 2x^3 - 2x^2 + 3x - 3
Now, differentiate each term with respect to x:
dy/dx = 6x^2 - 4x + 3
Now, to find the slope of the tangent line at x = 2, substitute x = 2 into the derivative equation:
dy/dx = 6(2)^2 - 4(2) + 3 = 24 - 8 + 3 = 19
Therefore, the slope of the tangent line to the graph of the equation y = (2x^2+3)(x-1) at x = 2 is 19 (option c).