Here are the four possible combinations for the parity of the degree and the sign of the leading coefficient:
If the lead coefficient is positive and the degree of polyonimial is odd, then the limits will be
[tex]\displaystyle \lim_{x\to-\infty}p(x)=-\infty,\quad \lim_{x\to\infty}p(x)=\infty[/tex]
In fact, odd powers maintain the sign (for example, [tex](-2)^3=-8[/tex] and [tex]3^3=27[/tex]. So, a very large number with an odd exponent will be even larger, and will keep the same sign.
In this case, everything we said about the previous case still holds, but the negative coefficient swaps all the sign. So, we have
[tex]\displaystyle \lim_{x\to-\infty}p(x)=\infty,\quad \lim_{x\to\infty}p(x)=-\infty[/tex]
Even exponents always return a positive value: if the input was positive the result remains positive (e.g. [tex]3^2=9[/tex]), and if the input was negative, the result becomes positive (e.g. [tex](-2)^4=+16[/tex]).
So, it doesn't matter if [tex]x[/tex] gets bigger and bigger in the positive or negative direction, the result will be positive nevertheless:
[tex]\displaystyle \lim_{x\to-\infty}p(x)=\infty,\quad \lim_{x\to\infty}p(x)=\infty[/tex]
Obviosuly, in this case, the negative leading coefficient causes all the signs to swap:
[tex]\displaystyle \lim_{x\to-\infty}p(x)=-\infty,\quad \lim_{x\to\infty}p(x)=-\infty[/tex]
PS: Throughout all the answer I only looked at the leading coefficient, because it's the one that decides the end behaviour of the polynomial. In fact, we have
[tex]\displaystyle \lim_{x\to\pm\infty}p(x)=\lim_{x\to\pm\infty} a_nx^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x+a_0[/tex]
If we factor the leading power we have
[tex]\displaystyle\lim_{x\to\pm\infty} x^n\left(a_n+\dfrac{a_{n-1}}{x}+\ldots+\dfrac{a_2}{x^{n-2}}+\dfrac{a_1}{x^{n-1}}+\dfrac{a_0}{x^n}\right)[/tex]
Since [tex]x[/tex] is going to infinity, all the terms with [tex]x[/tex] at the denominator will vanish in the limit, and so we have
[tex]\displaystyle\lim_{x\to\pm\infty} x^n\left(a_n+\dfrac{a_{n-1}}{x}+\ldots+\dfrac{a_2}{x^{n-2}}+\dfrac{a_1}{x^{n-1}}+\dfrac{a_0}{x^n}\right)=\displaystyle\lim_{x\to\pm\infty} a_nx^n[/tex]
And this limit depends only on the sign of the leading coefficient and the parity of the exponent, as we discussed above.
Answer:
C. degree 5, leading coefficient 1
on edge its right 2020
Step-by-step explanation:
