SOLUTION: 1. Given f(x)=2x^3+x^2-8x-4. (a) Find the x-intercepts of f. (b) Find f(0). (c) Use test points within the intervals formed by the x-intercepts to determine the sign of f(x)

Algebra ->  Graphs -> SOLUTION: 1. Given f(x)=2x^3+x^2-8x-4. (a) Find the x-intercepts of f. (b) Find f(0). (c) Use test points within the intervals formed by the x-intercepts to determine the sign of f(x)      Log On


   

Question 753830: 1. Given f(x)=2x^3+x^2-8x-4.
(a) Find the x-intercepts of f.
(b) Find f(0).
(c) Use test points within the intervals formed by the x-intercepts to determine the sign of f(x) in the interval.
(d) Graph f(x).

Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Most of the work is in (a) and (c). The x-intercepts are the zeros of the function, which you find using Rational Roots Theorem and synthetic division. The possible zeros to test are plus and minus 4, 2, and 1. Note that when you find any one of the zeros, you can factor or evaluate the quadratic quotient easily. You have six possible zeros, but you stand a good chance of needing not more than four synthetic divisions.

Testing root of +2:
+2____|___2___1___-8____-4
______|
______|_______4___10____4
___________________________
__________2___5___2_____0 _____NO REMAINDER

So factor into %28x-2%29%28x%5E2%2B5x%2B2%29
The quadratic expression factor is not factorable so use formula for quadratic solution:

The other two zeros are found to be %28-5-sqrt%2817%29%29%2F2 and %28-5%2Bsqrt%2817%29%29%2F2
.... unless I made any mistakes in the way. As decimal approximations, these two zeros are -4.56 and -0.438.

Your intervals to check are x%3C-4.56, -4.56%3Cx%3C-0.438, -0.438%3Cx%3C2, and 2%3Cx.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
1. Given f(x)=2x^3+x^2-8x-4.
(a) Find the x-intercepts of f.
use rational roots theorem to solve:
....0...|.....2......1......-8......-4....
....1...|.....2......3......-5......-9...
....2...|.....2......5.......2........0 (2 is a root)
f(x)=(x-2)(2x^2+5x+2)
f(x)=(x-1)(2x+1)(x+2)
roots are: 1, -1/2, -2 (x-intercepts)
..
(b) Find f(0).
f(0)=-4
..
(c) Use test points within the intervals formed by the x-intercepts to determine the sign of f(x) in the interval.
number line:
<...-...-2...+....-1/2...-....1....+.....>
(d) Graph f(x).
+graph%28+300%2C+300%2C+-4%2C+4%2C+-10%2C+10%2C2x%5E3%2Bx%5E2-8x-4%29+