r(x)= x^3+8/x^2+4
2006-11-18 07:27:07 · 5 answers · asked by jere 1 in Science & Mathematics ➔ Mathematics
r(x)= x^3+8/x^2+4
Assuming you mean r(x)= (x^3 + 8)/(x^2 + 4)
= (x + 2)(x^2 - 2x + 4)/(x^2 + 4)
When x = 0, r(x) = 8/4 = 2.
So it cuts the y- axis at (0, 2)
r(x) = 0 when (x + 2)(x^2 - 2x + 4)/(x^2 + 4) = 0, ie when x = -2
So it cuts the x-axis at (-2, 0)
ie x-intercept is -2 and y-intercept is 2.
2006-11-18 08:27:18 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
let r(x)=y
then y-intercept =>x =0
r(0) = 4
x-intercept => y=0
x^3 +8/x^2 +4=0
2006-11-18 15:34:36 · answer #2 · answered by anami 3 · 0⤊ 0⤋
f(x)=x^3+8/(x^2+4
setting x=0 y intercept=(0,2)
settingy=0
x^3+8=0
(x+2)(x^2-2x+4)=0
x=-2 or imaginary
x intercept is (-2,0)
2006-11-18 15:41:25 · answer #3 · answered by raj 7 · 0⤊ 0⤋
put x=0 to get y intecept
and y=0 to get x intercept
2006-11-18 15:30:14 · answer #4 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋
http://froogle.google.com/froogle?q=r(x)%3D+x%5E3%2B8/x%5E2%2B4&hl=en&lr=&sa=X&oi=froogle&ct=title
2006-11-18 15:29:56 · answer #5 · answered by god knows and sees else Yahoo 6 · 0⤊ 0⤋