how do i turn x^3-6x^2+11x-6 into a quadratic equation?

Question

Well the question asks me to sketch the graph, but in order to do that i need to find the other points right? so i need to find the quadratic equation? so it would be ... x(x^2-6x+11 .. and what do i do with the 6?

help :(

Answer 1

alright champ, here's the deal:

x^3-6x^2 + 11x-6 <-- this is of degree 3 -- a cubic expression.
that's the other thing, what you give is an expression...

i assume you're going to be graphing y = x^3-6x^2 + 11x-6.

well, being a cubic and all, you can't just magically 'find the quadratic equation,' unfortunately.

it can still be graphed though, but graphing it by hand is lame.
i think that is typically an 'algebra 2' topic. have fun learning how to do it, you'll never need to do it again a few months from now i bet.

here are some resources:

http://www.google.com/search?hl=en&q=graphing+cubic+polynomials

http://mathforum.org/library/drmath/view/52633.html

http://mathforum.org/dr.math/faq/faq.cubic.equations.html

check those out, particularly the google search.

there are some neat interactive examples to play around with [like embedded applets] that will let you get a better feel for how cubic polynomials behave.

enjoy

Answer 2

Hi I think you are a little confused. You have a cubic equation you can not turn a cubic equation into a quadratic equation. I think you are asking how do you graph the cubic equation. So what you need to do is factor the equation. To do this I recommend you use long division. The tutorial below is pretty good. The end result is (x-1)(x-2)(x-3) = 0 These points tell you where the function crosses the x-axis. You also have the y-intercept point (0,-6). To figure out the height of the bumps solve for x = 1.5 and x = 2.5

Answer 3

u can use the remainder theorem

f(1)= 1^3-6(1)^2+11(1)-6=0
hence (x-1) is a factor

x^3-6x^2+11x-6=(x-1)(x^2+kx+6)
by comparing coeffients of x^1,
11=k+6
k=5

x^3-6x^2+11x-6=(x-1)(x^2+5x-6)
=(x-1)(x-2)(x-3)

hence u get the roots, 1, 2 and 3

if you want to fing the y-intercept, substitute x=0 and u get -6
i suppose you can sketch ur graph now(:

Answer 4

the general rule to solve that just put x=1, -1, 2 and so on. the value of x for which the expression is 0 is one of d factors of d expression.
here we can see that for x=1 it is 0.hence x^3-6x^2+11x-6 can be written as (x-1)(x^2-5x+6)
to obtain the 2nd term i have just divided d whole expression by x-1
we have reduced d expression to
(x-1)(x^2-5x+6)
= (x-1)(x^2-3x-2x+6)
=(x-1)(x(x-3)-2(x-3))
=(x-1)(x-2)(x-3)
now i can assume
dat u can draw graph

Answer 5

6x² - 5x = 6 6x² - 5x - 6 = 0 .............. seem for factors of 6•6 that fluctuate through 5 ? 9&4 6x² - 9x + 4x - 6 = 0 3x(2x - 3) + 2(2x - 3) = 0 (3x + 2)(2x - 3) = 0 x = -2/3 or x = 3/2