Question

Can someone help me with finding the points on the graph at which the tangent line has a slope of 1.
y=1/3x^3-2x^2+4x+1

Answer 1

I know I have to take the derivative of this. Which i believe will be Y'=x^2-4x+4

Answer 2

correct, and that formula there defines the slope of the tangent line at any given value of x

what would you spose we do with it if we want to know which points have a tangent line slope of 1?

Answer 3

set it equal with a slope of 1?

Answer 4

yep, sooo

1=x^2-4x+4 is a quadratic equation from algebra ... can you find the values of x for this?

Answer 5

or if you notice the complete squareness of the right side ...

1 = (x-2)^2 ; when x = +-1 + 2 , correct?

Answer 6

yes sorry I'm a little slow I have to right everything out. but yes i got the same thing (x-2)(x-2)=1 and then splitting it up i got x=2 but i didn't get x=-1

Answer 7

when x=2, y' = 0

in order to adjust the value of x, such that y' = 1 .... x = 2-1, or x = 2+1

Answer 8

oh okay i see

Answer 9

now that we know for what values of x we get a slope of "1" ... x = 1 or x=3

lets find our y values for the points they want.

y = f(x); determine the value of f(1) and f(3) to define the points: (1,f(1)) and (3,f(3))

Answer 10

so pluging in the x values we get y'=1^2+4(1)+4 which makes y=1and y'=3^2-4(3)+4 which would also make y= 1 but i don't think that's right

Answer 11

or do i put it into the original equation? and not y'?

Answer 12

the original equation defines points

the derivative of the equation defines slopes

so we determined the values of x to define a slope of 1, now we would want to use the original to define the points where the slope is 1

Answer 13

oh okay so its going to be (1,10/3) for the girst point and for the second you get (3,4)

Answer 14

ill trust you on those ;)

Answer 15

i ment first ha sorry.

Answer 16

okay thanks so much i really appriciate it

Answer 17

good luck ;)

Answer 18

thanks so much