2 questions - applicable differenation and intersection

[klli]

Homework Statement

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

 

[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)

 

[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 I am 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.

 

[klli]

thanks man appericiate it