Solve the given system by substitution.
2 answers:
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Your problem statement makes no reference to p or q, so we can only guess that "p/q" refers to the ratio of the lowest- and highest-degree coefficients:
... -6/1
Then the factors of p/q are the factors of 6, with either sign.
Factors of p/q: ±1, ±2, ±3.
There are a number of ways the factors can be tested. To test 1, simply add the coefficients. (The sum is -8.) To test -1, subtract the sum of odd-degree coefficients from the sum of even-degree coefficients. (The difference is 0.)
Knowing that -1 is a root of the function, you can use synthetic division to find the quadratic factor. (See the first attachment for the work.) Then you have ...
... f(x) = (x +1)(x² +x -6)
You can continue to test the factors of p/q, or you can factor the quadratic to get ...
... f(x) = (x +1)(x -2)(x +3)
Another way to test the factors is to program your graphing calculator to evaluate the function f(x) for those factors. The result from my TI-84 is shown in the second attachment. That shows f(x) = 0 for x = -3, -1, and +2.
Test the factors: f({-3, -2, -1, 1, 2, 3}) = {0, 4, 0, -8, 0, 24}
This tells you the x-intercepts.
X-intercepts: -3, -1, 2
Of course, the y-intercept is just the constant in the function—the value of the function when x=0.
Y-intercept: -6
As with any odd-degree polynomial with a positive leading coefficient, the overall behavior looks like "/". That is ...
End behaviors: f(x) goes to -∞ as x goes to -∞; f(x) goes to +∞ as x goes to +∞.
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A graph of f(x) is shown in the third attachment. (Graphing calculators make it easy to use graphing to find x-intercepts.)
