Find the horizontal and vertical asymptotes of the graph of the function f(x) = 2x^2+1/3x-5

HA= None VA= 5/3

Step-by-step explanation:

The HA is none because the bottom part is missing x squared so you add x squared and it will be 2/0 which means it's not possible which means HA= none. VA is 5/3 because you set the denominator to 0 so you add 5 then divide by 3 so the answer for VA is 5/3

do you have any choices

step-by-step explanation:

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)