Question

f(x)=-2x^3+6x^2-3 how do i find the x intercepts?

Answer 1

I tried factoring it and it ends up -2x^2(x-3) and 3 isnt an x intercept.

Answer 2

the x intercepts are a negative fraction and a positive fraction. Ive already graphed it im just trying to find out the long way.

Answer 3

whhoops. -2(x-3)x^2

Answer 4

\[f(x) = -2x^3+6x^2-3\] and find when \[f(x) = 0\]You can use the cubic formula for difficult cubics. (it's a bit too long to memorise) but I think that might be your only option!
You can find it http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Answer 5

Oh lord. yea. i had it set equal to 0. i tried doing it 2x^3=6x^2-3
also as 3=-2x^2+6x^2

Answer 6

neither seem to work.

Answer 7

Other alternate forms are \[(6-2x)x^2-3\]and\[-2(x-2)^3-6(x-2)^2+5\]

Answer 8

thank you :)
Miscalculated.

Answer 9

can you do an example of this equation? @_@

Answer 10

Er, I'll give it a go. give me a minute

Answer 11

Sure lol. thanks. Ive never done the cubic formula it looks like itll end up with a sqrt of something but kind of boggling my mind right now.

Answer 12

Yeah, it's mega complicated, and can lead to complex numbers. (and does in your case too) however it should pop out with 3 numbers. With your formula\[a = -2\]\[b=6\]\[c=0\]\[d=-3\]Substituting that into crazy equation gives...(this might take a while)

Answer 13

... really?
im getting x intercepts of:
- 0.64178
0.83174
2.81

Answer 14

wow, thats correct!

Answer 15

i think hea found my y intercept but the x's were incorrect.

Answer 16

are we sure the equations right, that's a hell of a weird group of intercepts, usually they make these answers a bit cleaner

Answer 17

yes. The original equation is f(x)=-2x^3+6x^2-3

Answer 18

the first deriative is -6x^2+12x
and the second is -12x+12

Answer 19

my critical numbers were x=0 and x=2. min is -3 and max is at 5.
seems to all work out?

Answer 20

unless i derivative'd wrong. lol

Answer 21

My math teacher is a jerk.

Answer 22

I'm a little lost now, but basically \[-2x^3+6x^2-3=0\]The solutions are \[x=-0.642\]\[x=2.810\]Does that sort it?

Answer 23

yea, but howd you get the x intercepts? D:

Answer 24

the cubic?

Answer 25

Yeah subbing the values from above into the mega long equation.

Answer 26

@hea you missed one hea, if you graph it: it passes through the x axis 3 times

Answer 27

Thanks. This is a graphing question and im only in Calc 1. my teacher is sort of vague on these things but im assuming we are meant to graph it and find it from the graph then. its just weird because usually we do it the longhand way. the 0=-2x^3+6x^2-3 and solving for x. Do you guys think he meant for us to use the cubic formula?

Answer 28

True my bad,\[x=0.832\]However I missed this because the solution is complex, however it is of course still valid

Answer 29

Probably not, yeah I reckon teach meant for you to graph it and then estimate the result from that. Look in to various root finding methods, Newton's method, Bisection method and on and on. Is the kind of thing that your teacher has been going on about?

Answer 30

nope. we had another question similar to this and he expects us to just estimate. thanks for all the help :)

Answer 31

id get rid of the x^2 term, like cardano

Answer 32

2x^3-6x^2+3 = 0

x^3-3x^2+3/2 = 0 ; let x = (u+1)

(u+1)^3-(u+1)^2+3/2 = 0

u^3+u^2+u+1
-u^2-2u-1
+3/2
---------------
u^3 -u+3/2 = 0


u^3 = u - 3/2

Answer 33

then there is a cubic formula that is derived .... similar to the quadratic formula