Respuesta :
Hello there. To solve this question, we have to remember some properties about polynomial functions.
Given the polynomial function
[tex]p(x)=3(x+1)(x-2)(2x-5)[/tex]We want to determine:
a) What are the x-intercepts of the graph of p(x)?
For this, we have to determine the roots of the polynomial function p(x). In this case, we have to determine for which values of x we have
[tex]p(x)=0[/tex]Since p(x) is written in canonical form, we find that
[tex]p(x)=3(x+1)(x-2)(2x-5)=0[/tex]A product is equal to zero if at least one of its factors is equal to zero, hence
[tex]x+1=0\text{ or }x-2=0\text{ or }2x-5=0[/tex]Solving the equations, we find that
[tex]x=-1\text{ or }x=2\text{ or }x=\dfrac{5}{2}[/tex]Are the solutions of the polynomial equation and therefore the x-intercepts of p(x).
b) What is the end-behavior of p(x) as x goes to +∞ or x goes to -∞?
For this, we have to take the limit of the function.
In general, for polynomial functions, those limits are either equal to ∞ or -∞, depending on the degree of the polynomial and the leading coefficient.
For example, a second degree polynomial function with positive leading coefficient is a parabola concave up and both limits for the function as x goes to ∞ or x goes to -∞ is equal to ∞.
On the other hand, an odd degree function usually has an odd number of factors (the number of x-intercepts in the complex plane) hence the limits might be different.
In this case, we have a third degree polynomial equation and we find that, as the leading coefficient is positive and all the other factors are monoic, that
[tex]\begin{gathered} \lim_{x\to\infty}p(x)=\infty \\ \\ \lim_{x\to-\infty}p(x)=-\infty \end{gathered}[/tex]That is, it gets larger and larger when x is increasing arbitrarily, while it get smaller and smaller as x is decreasing.
c) To find the equation for a polynomial q(x) that has x-intercepts at -2, 3/4 and 7.
The canonical form of a polynomial of degree n with x-intercepts at x1, x2, ..., xn and leading coefficient equals a is written as
[tex]f(x)=a\cdot(x-x_1)(x-x_2)\cdots(x-x_n)[/tex]So in this case, there are infinitely many polynomials satisfying this condition. Choosing a = 1, we find that q(x) is equal to
[tex]\begin{gathered} q(x)=(x-(-2))\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7) \\ \\ \boxed{q(x)=(x+2)\cdot\left(x-\dfrac{3}{4}\right)\cdot(x-7)} \end{gathered}[/tex]These are the answers to this question.
