Question

A polynomial function g(x) has a negative leading coefficient. Certain values of g(x) are given in the following table. x –4 –1 0 1 5 8 12 g(x) 0 3 7 12 4 3 0 If every x-intercept of g(x) is shown in the table and each has a multiplicity of one, what is the end behavior of g(x)?

Answer 1

Using the factor theorem and limits, it is found that the end behavior of g(x) is that it goes to negative infinity.

-------------------------------

The Factor Theorem states that a function with leading coefficient a and zeros $x_1, x_2, ..., x_n$ is defined by:

$g(x) = a(x - x_1)(x - x_2)...(x - x_n)$

In the given table, $f(x) = 0$ for $x = -4, x = 12$, thus the zeros are: $x_1 = -4, x_2 = 12$

Then, the definition is:

$g(x) = a(x - (-4))(x - 12)$

$g(x) = a(x + 4)(x - 12)$

$g(x) = a(x^2 - 8x - 48)$

The end behavior is the limit of g(x) as x goes to infinity, thus:

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} ax^2 = a \times \infty$

Since the leading coefficient a is negative:

$a \times \infty = -\infty$

Thus, the end behavior of g(x) is that it goes to negative infinity.

A similar problem is given at https://brainly.com/question/24248193

Answer 2

Answer:

As x→–∞, g(x)→–∞ and as x→∞, g(x)→–∞.