# Homework Help: 2 questions - applicable differenation and intersection

1. Aug 1, 2008

### klli

1. The problem statement, all variables and given/known data

hey guys ok basically i have 2 questions, first one is

(1) "find the point on the curve y=(x-1)[power of 5] +2 where the tangent to the curve at these points is parallel to the x axis"

Attempt : well the gradiet of the x axis is 0. So i differentiated the equation and i got myself
y'=5(x-1)[power of 4] +2 and then i substutitued 0 into y and then used the solver funnction in my calculator but its giving me x = 101.something

The answer is ment to be (1,2)

(2) for this question i just want to know the general process of how i would find the points of intersection between a quadratic and a linear equation , as well between 2 quadratic equations. ( you are provided the equation of the curve/line)

thanks

2. Aug 1, 2008

### physicsnoob93

Ok for the first one,

Remember, when you differentiate a constant it will give you 0. So your y' = 5((x-1)^4)
Let y' = 0
0 = 5((x-1)^4))
(x-1)^4 = 0.
x-1 = 0.
x = 1
So sub back 1 into y = (x-1)^5 + 2
You'll get y = 2.
So the point is (1,2)

3. Aug 1, 2008

### physicsnoob93

For the second one, Since the 2 equations intersect, i can say that there is a point P(x1,y1) such that it exists and is the same for both equations.

So if im given the equation for both the functions, i'll just sub in x1 into both. And the value of the function for both of them would be y1, and i can equate them.
And then solve for x1.

But in your case, since you want to find the point of intersection, you would first have to propose a certain point which is the same for both functions before you sub them in.

4. Aug 1, 2008

### klli

thanks man appericiate it