x^3 + x^2 - 8x + 2 = -7x + 3
x^3 + x^2 -x - 1 = 0
x^2(x+1) - (x+1) = 0
(x^2-1)(x+1) = 0
you shoud be able to finish now
Find all the points where the tangent to the curve at (-1,10) meets the curve.
First, I differentiate the curve and then find the slope of the tangent for when x=-1.
And I found the equation of this tangent to be , although I'm not sure what to do with this exactly.
The only other thing I've been able to deduce is that the slope of the tangent of any other points hitting the curve at (-1,10) should also be -7. I have a few more questions in my homework similar to this one but I'm not sure of the method for doing them. Help please!
The normal to the curve , where a and b are constants, has equation at the point x =4. Find the values of a and b.
So first I find the value of the tangent at x=4.
-4 is the slope of the normal (*i think*) so the slope of the tangent is 1/4.
Next I use this slope in the derivative of the curve.
I know I'm aiming for two equations in the end so that I can use elimination or the two equations/two unknowns method for solving for a and b, but I'm getting lost with the second equation here.
If I solve for y by subbing in x=4, do I get the value of the tangent line or the normal line? I tried doing it through with both but I don't get a clear answer for either. In both cases, I end up with and a second equation where both my a and b end up cancelling each other out so that I'm left with 0. What am I doing wrong?
Thank you for reading, help is appreciated !