Given: f(x)= x^3 + x^2 - 2x over the closed interval [0,2]

Write the equation of the tangent line to f(x) passing through point A(-1,-2).

*At first glance I knew how to do it, but f(-1) does not equal -2...so do I add a constant onto f(x) to make f(-1) = -2 or what? thanks

To write the equation of the tangent line to f(x) passing through point A(-1, -2), you don't need to modify the original function f(x). The discrepancy you noticed between f(-1) and -2 is simply because A(-1, -2) is not on the graph of f(x).

To find the equation of the tangent line, we need two pieces of information: the slope of the tangent line and a point on the line.

1. Finding the slope of the tangent line:
The slope of the tangent line is equal to the derivative of the function evaluated at the point where the tangent line touches the curve. Let's find the derivative of f(x) first:

f(x) = x^3 + x^2 - 2x

Differentiating f(x) with respect to x, we get:
f'(x) = 3x^2 + 2x - 2

Now, we can evaluate the derivative f'(x) at the x-coordinate of the point A(-1, -2):
f'(-1) = 3(-1)^2 + 2(-1) - 2
= 3 + (-2) - 2
= -1

So, the slope of the tangent line is -1.

2. Finding a point on the tangent line:
We already have the point A(-1, -2).

Now that we have the slope (-1) and a point (-1, -2), we can use the point-slope form of a linear equation to write the equation of the tangent line:

y - y1 = m(x - x1)

Using the slope (-1) and the point A(-1, -2), we substitute the values into the equation:

y - (-2) = -1(x - (-1))
y + 2 = -(x + 1)
y + 2 = -x - 1
y = -x - 3

Thus, the equation of the tangent line to f(x) passing through point A(-1, -2) is y = -x - 3.