Select the function that represents a parabola with vertex at (2,1) and y-intercept (0,3). Question 9 options: A) ƒ(x) = 1∕2(x + 2)2 + 1 B) ƒ(x) = 2(x – 2)2 + 1 C) ƒ(x) = 1∕2(x – 2)2 + 1 D) ƒ(x) = 2(x + 2)2 + 1

answer: need picture of problem

step-by-step explanation:

the answer is d

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.

the answer should be:

a. 3, 4, and 6

b. 18, 24, and 36

c. 4.8, 6.4, and 9.6

hope i

-lavira : )