Question

How to find the point of f(x)= (3x-2x^2)^2 where the tangent line is parallel to the x-axis?

Answer 1

firtst of all you need to find then slope of the tangent by dfferentiating f(x), do you know how to do that?

Answer 2

or what you could do is factorize f(x)

Answer 3

for a line to be parallel to the x axis then x=0

Answer 4

so plug zeros in the x's of f(x)

Answer 5

y=? that would be parallel to the x-axis and this is the third method you can use...

Answer 6

i used to product rule and got f' (x) to be 16x^3 -12x^2-6x. I don't know if its right thought, because the answers are (0,0) (3/2, 0) and (3/4, 81/64)

Answer 7

no thats wrong

Answer 8

ok. what would it be though?

Answer 9

\[f(x)= (3x + -2x^2)^2\]

Answer 10

the best way with less error to solve this function is to expand it...

Answer 11

then differentiate it

Answer 12

\[(a+b)^2 = a^2 +2ab+b^2\]

Answer 13

(3x-2x^2)(3x-2x^2). But when I used the product rule to differentiate it, i got the above equation

Answer 14

well i would do it another way

\[f(x)= (3x + -2x^2)^2 = 9x^2-6x^3+4x^4\]

Answer 15

now you can differentiate it simply

Answer 16

then would i use the power rule to differentiate it

Answer 17

yes

Answer 18

so i would get 16x^3-18x^2+18x

Answer 19

yes

Answer 20

but i don't get how to get the other two points as the answer. I get that if i sub 0 in, then i will get 0. but I don't get how to get (3/2, 0) and (3/4, 81/64)

Answer 21

you need to factor the third order polynomial...

Answer 22

can I long divide it, if i find an x value that makes it equal to 0

Answer 23

16x^3-18x^2+18x= 2x(8x^2-9x+9)

Answer 24

i'm sorry, but i still don't know how to get those values

Answer 25

well you use the quaratic equation

Answer 26

from what i can see you are going to get a complex number, have you done that before...

Answer 27

i tried, i get a negative under the square root

Answer 28

yes that is a complex number...

Answer 29

ok do you know the chain rule

Answer 30

no

Answer 31

then you need to check if the function is correct

Answer 32

ok . I guess i will just ask my teacher

Answer 33

ok find the 2nd derivative

Answer 34

what?

Answer 35

i didn't learn that yet

Answer 36

ok there is a simpler way

Answer 37

i will just ask my teacher, it ok. thanks

Answer 38

well you know that x=0, right the one we factored out

Answer 39

so we said when x =0 that will give you the value of y, and we get y=0

Answer 40

so the line passing through the piont(0,0) is the line that is parallel to the x-axis

Answer 41

based on basic geomatry thoerem

Answer 42

i don't know what that is

Answer 43

as i said at the beginig that when x=0 y=?

Answer 44

y = 0

Answer 45

ok let me put it this way

set y =0, what values for x would you get?

Answer 46

(3x-2x^2)^2 = 0 -> x (3-2x)=0

Answer 47

what is x =?

Answer 48

3/2

Answer 49

so we get the point (3/2, 0)

Answer 50

did u get that?

Answer 51

\[\frac{y-y_1}{x-x_1}=m\]

Answer 52

yea

Answer 53

ok

Answer 54

now remeber a line parallel to an axis tells you that the slop is =0 because there is no slope right, its a straight line...

Answer 55

yes

Answer 56

do you know how to complete the square

Answer 57

because when you complete the square you'll find the vertix which would give you the point parallel to the x axis

Answer 58

yes

Answer 59

ok good find the square and i will see you later because its too late here 5am in the morning ...

Answer 60

take care and good lcuk

Answer 61

ok, thank you