y = x3 - 4x2 + 1x + 6 This example has y intercept (0, 6) Since the first term is positive, just like our introductory example, the graph will come from negative infinity on the left, and travel upwards to the right, to positive infinity. In order to find the x intercepts, we'll have to factor x3 - 4x2 + 1x + 6 The Integral Zero Theorem tells us that, because the first coefficient is 1, there will be at least one zero that is an integer, and it will be a positive or negative factor of the constant term 6 The factors of 6 are +/- 1, 2, 3, 6 When we find the value a that is a zero; ie: makes the polynomial have a value of zero when substituted for x, the Factor Theorem tells us that we will have one factor (x - a) of the polynomial. We need to test each one. Testing +1: x3 - 4x2 + 1x + 6 = 13 - 4(1)2 + 1(1) + 6 = 4 so 1 is not a zero. Testing -1: x3 - 4x2 + 1x + 6 = (-1)3 -4(-1)2 +1(-1) + 6 = 0 so -1 is a zero. We now know that (x - -1) or (x + 1) is one factor of x3 - 4x2 + 1x + 6 If we were to divide the original polynomial by (x + 1), we would be left with a quadratic polynomial, which we know how to solve, either by factoring or using the general quadratic formula. So we'll use (synthetic) division:(we won't show the work) (x3 - 4x2 + 1x + 6) ÷ (x + 1) = x2 - 5x + 6 and x2 - 5x + 6 factors into (x - 3)(x - 2) We now have all three factors of the original polynomial: x3 - 4x2 + 1x + 6 = (x + 1)(x - 3)(x - 2) So the three zeros of the polynomial are -1, 3, and 2 The x intercepts are (-1, 0), (3, 0) and (2, 0) 
We have points (0, 6) (y intercept) and (-1, 0), (3, 0) and (2, 0) (x intercepts) Knowing the end behaviours, this is all we need to draw the graph. Remember that we don't know the heights of the loops. Next: what happens when two or three x intercepts are the same. |