Determine the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2)

dy/dx= 6(x^2+x-1)^2 (2x+1)

at (-1,-2)

dy/dx= 6(1-1-1)^2 (-2+1)
= -6

so now you have the slope of -6 and a point(-1,-2)

continue ....

So

y2-y1=m(x2-x1)

y-(-2)=-6(x-(-1)
y=6x-1

therefore the slope of tangent is 6?

To determine the slope of the tangent to the curve at the given point, you can use the derivative.

The derivative of a function gives you the slope of the tangent line at any point on the curve. In this case, we need to find the derivative of the function y=2(x^2+x-1)^3.

To find the derivative, we can use the chain rule. The chain rule states that if you have a function of a function (in this case, a cubic function of another function), you need to multiply the derivative of the outer function by the derivative of the inner function.

Let's find the derivative step by step:

Step 1: Start by using the power rule. Multiply the exponent by the coefficient and subtract 1 from the exponent:
d/dx [ (x^2+x-1)^3 ] = 3(x^2+x-1)^2 * d/dx [ x^2+x-1 ]

Step 2: Now, find the derivative of the inner function:
d/dx [ x^2+x-1 ] = 2x + 1

Step 3: Substitute the derivative of the inner function into the derivative of the outer function:
d/dx [ (x^2+x-1)^3 ] = 3(x^2+x-1)^2 * (2x + 1)

Now we have the derivative of the function y=2(x^2+x-1)^3.

To find the slope of the tangent line at (-1, -2), we can substitute x=-1 into the derivative:

slope = 3((-1)^2 + (-1) - 1)^2 * (2(-1) + 1)

Simplify the expression:

slope = 3(1 - 1 - 1)^2 * ( -2 + 1)
slope = 3(-1)^2 * (-1)
slope = 3 * -1 * -1
slope = 3

Therefore, the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2) is 3.