Hi, I'm absolutely stuck on this problem. This is for a newton's method/linearization unit. Find an equation for the curve in the xy-plane that passes through the point (1,-1) if its slope at x is always 3x^2+2.
dy/dx = 3x^2+2
so
y = x^3 + 2 x + c
if x = 1, y = -1
-1 = 1 + 2 + c
c = -4
y = x^3 + 2 x - 4
I do not know why you would want to use Newton's Method.
This was on a test I took earlier and I'm going back now to fix my mistakes. Thank you Damon-you've been a big help!
To find an equation for the curve in the xy-plane that passes through the point (1, -1) and has a slope of 3x^2 + 2 at every x value, we can use the process of integration.
The first step is to integrate the given slope equation with respect to x. Integrating 3x^2 + 2, we get:
∫(3x^2 + 2) dx = x^3 + 2x + C
Where C is the constant of integration.
Next, we need to substitute the coordinates of the given point (1, -1) into the integrated equation to find the specific value of C.
When x = 1, the equation becomes:
(1^3) + 2(1) + C = -1
1 + 2 + C = -1
3 + C = -1
C = -4
Now we have the specific value of C, which is -4. We can substitute this value back into the integrated equation to obtain the equation for the curve:
x^3 + 2x - 4 = y
So, the equation for the curve in the xy-plane that passes through the point (1, -1) and has a slope of 3x^2 + 2 at every x value is y = x^3 + 2x - 4.