Find the third degree polynomial function that has an output of 40 when x = 1, and has zeros - 19 and -i?

f(x) = x³ + 19x² + x + 19

Step-by-step explanation:

Zeros: -19 , -i, complex conjugate root: +i

Function: (x+19)(x+i)(x-i) = 0

(x + 19) (x² - i²) = (x + 19) ( x² + 1) = x³ + 19x² + x + 19

check: when x = 1

f(1) = 1 + 19 + 1 +19 = 40

multiply then divide

step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.