Determine the slope of the tangent line to the curve y=(x^2+3x)(x^2-2) at x=-1.
I got 3 as my answer although I am not too sure that it's correct. I found the derivative of the function and then substituted -1 into the equation.
that's the proper method.
Using the product rule,
y' = (2x+3)(x^2-2) + (x^2+3x)(2x)
at x=-1, that's (-2+3)(1-2)+(1-3)(-2) = 3
I do not have the same value as yours.
Can you post the derivative?
Oh, sorry, I had the two terms divided and not multiplied.
thanks!
To determine the slope of the tangent line to the curve at a specific point, you need to find the derivative of the function and evaluate it at that point. Here's how you can do it step by step:
1. Start by finding the derivative of the function y=(x^2+3x)(x^2-2). To do this, you can use the product rule, which states:
If y = f(x)g(x), then y' = f'(x)g(x) + f(x)g'(x).
Applying the product rule, differentiate each term separately:
f(x) = x^2+3x, so f'(x) = 2x + 3
g(x) = x^2-2, so g'(x) = 2x
Now, substitute these values into the product rule:
y' = (2x + 3)(x^2-2) + (x^2+3x)(2x)
2. Simplify the expression obtained in step 1. Expand and combine like terms:
y' = 2x^3 - 4x + 3x^2 - 6 + 2x^3 + 6x^2
= 4x^3 + 9x^2 - 4x - 6
3. Now that you have the derivative, you can find the slope of the tangent line by evaluating it at x = -1. Substitute -1 into the derivative expression:
slope = 4(-1)^3 + 9(-1)^2 - 4(-1) - 6
= 4(-1) + 9(1) + 4 - 6
= -4 + 9 + 4 - 6
= 3
Therefore, the slope of the tangent line to the curve y=(x^2+3x)(x^2-2) at x=-1 is 3. Your answer is correct!