The characteristics of the graph depends on the leading coefficient and
the degree of the polynomial.
Responses:
a. From the positive leading polynomial of the odd degree polynomial;
- The right end of the graph points upwards, and the left end of the graph will points downwards
b. x = 4, x = 0.5, and x = -3
c. Yes the graph crosses the x-axis
d. The y-intercept is the point y = 12
e. 2 turning points
f. Additional points of the graph are(1, -12), (3, -30) and (5, 72)
g. Please find attached the graph of the function.
Which properties determines the behavior of a polynomial?
The given function is; P(x) = 2·x³ - 3·x² - 23·x + 12
a. Given that the leading coefficient is positive and the degree is odd, we have;
- The direction in which the right end of the graph will point will be upwards, and the direction in which the left end of the graph will point is downward
b. The zeros of the function are found as follows;
P(x) = 2·x³ - 3·x² - 23·x + 12
At the zeros, 2·x³ - 3·x² - 23·x + 12 = 0
By factorization, using an online tool, we have;
2·x³ - 3·x² - 23·x + 12 = (x - 4)·(2·x² + 5·x - 3) = (x - 4)·(2·x - 1)·(x + 3) = 0
Alternatively, we have;
By trial and error, we have, at x = 4, 2·4³ - 3·4² - 23·4 + 12 = 0
Therefore;
(x - 4) is a factor
(2·x³ - 3·x² - 23·x + 12) ÷ (x - 4) = (2·x² + 5·x - 3)
(2·x² + 5·x - 3) = 2·x² + 6·x - x - 3 = 2·x·(x + 3) - 1(x + 3) = 0
(2·x² + 5·x - 3) = 2·x·(x + 3) - 1(x + 3) = (2·x - 1)·(x + 3)
Which gives;
2·x³ - 3·x² - 23·x + 12 = (x - 4)·(2·x - 1)·(x + 3)
Therefore, the zeros are;
x = 4, x = [tex]\frac{1}{2}[/tex] = 0.5, and x = -3
c. The maximum number of turning points in a cubic function are 2
Given that the graph has three zeros at x = -3, x = 0.5, and x = 4, the
graph crosses the x-axis to get three zeros from the two turning points.
d. The y-intercept is given by the point at which the graph crosses the y-axis, which is the point, x = 0 which is found as follows;
f(0) = 2·0³ - 3·0² - 23·0 + 12 = 12
- The y-intercept is the point on the graph with y = 12
Which gives the point (0, 12)
e. The number of turning points of a polynomial of degree n is n - 1
Therefore;
- The number of turning point of the polynomial 2·x³ - 3·x² - 23·x + 12 having 3 zeros are 2 turning points.
f. Additional points of the graph are;
f(1) = 2·1³ - 3·1² - 23·1 + 12 = -12
f(3) = 2·3³ - 3·3² - 23·3 + 12 = -30
f(5) = 2·5³ - 3·5² - 23·5 + 12 = 72
Therefore;
The points, (1, -12), (3, -30) and (5, 72) are points on the graph
g. Please find attached the graph of the function created with MS Excel
Learn more about the graphs of functions here:
https://brainly.com/question/14387154
