first i set up the problem, i put 1+2x-xcubed-1/x-1
i got stuck because of the xcubed, i forgot how to factor out polynomials that hae xcubed in them, i know this part doesn't have to do with calculus, but i forgot when i did polynomials like this
the answer isy=-x+3, but i need to figure out the slope
You've got the slope in this formula. y=-x+3 is the same as saying y=-1x+3. And since y=mx+b, what do you suppose m is?
I got the answer without factoring anything.
I'm not sure how you got the slope (even though you don't realize you got it), but what I did was take the derivative of 1+2x-x^{3} Then I plugged my x-value into that formula.
here is what the slope form looks like 1+2x-x3-1/x-1 which is f(x)-a/x-a
the only part i am having trouble with is the 1+2x-x3, so what do i do with this
When I plug in x=1 into the equation, the nominator and denominator both go to zero. So I need to use Hopital's rule and take the derivative of the top and the bottom.
When you have a fraction inside of a limit, and the numerator and denominator both go to zero or infinity, then you need to differentiate the numerator and the denominator. Just treat them separately though, don't go use the quotient rule.
Yes, that was the whole idea: the derivative of a function, at a given x, is the slope of the tangent line to the graph at that point! If you do not know that you cannot possibly do the problem. If do know that, then it is easy.
dy/dx = gradient of tangent = 2-3x squared
Then sub in point (1,2) and the gradient is -1
y=mx+b
we know m = -1
y= -x + b
Because the point (1,2) satisfy both the tangent and curve you sub it in
2= -1 + b
b= 3
therefore the equation of the tangent is y= -x + 3