y-int??? f(x)=3x^3-17x^2+18x+8
y=8 is y intercept
set x=0 and solve for y
(0,8) y intercept
3x^3-17x^2+18x+8=0
how do u find the x-int??
yes thats how yet set 3x^3-17x^2+18x+8=0 solve solve for x
x=2 works
i have stuck with x-int i dont know why i cant find it...
(x-4)(x-2)(3x+1)
2| 3 -17 18 8 | 6 -22 -8 ----------------- 3 -11 -4 | 0
:)..how did u get x=2??? it's right but i wonder know why??
you can guess using the rational root theorem... but there's a lot of guesses you can make for the root. It's somewhat trial and error unless you use calculus to determine the interval of one of the roots.
3x^2-11x-4=0 \[x=\frac{11 \pm \sqrt{121-4(3)(-4)}}{6}=\frac{11 \pm \sqrt{121+48}}{6}=\frac{11 \pm \sqrt{169}}{6}=\frac{11 \pm 13}{6}\]
right its a guessing thing
i mean not really guessing you have a set of rational numbers and you can do trial and error i just chose 2 and it worked
all the rational roots would be ±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3.
you look at factors of 8 and you look at factors or 3 and you also ook at factors of 8/3
and he listed them all
so we see we have two rational answers using the quad formula they should be in his list
we have x=24/6=4 and x=-2/6=-1/3
ok... thanks for everyone helping me!!!!
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