Consider the rational function: f(x) = 2x-5/3x+4

How do I work out the following with exact detailed steps?

Consider the rational function: f(x) = 2x-5/3x+4

Find the coordinates of the y-intercept of the graph of f(x).

Find the equations of both the horizontal and vertical asymptotes of f(x). State clearly which equation matches each asymptote.

Please (I've mentioned this too many times to count) use parentheses:

f(x) = (2x−5)/(3x+4)

y-intercept always occurs when x = 0
y = f(0) = (0−5)/(0+4) = −5/4

Coordinates of y-intercept: (0, −5/4)

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Vertical asymptote where denominator = 0
3x + 4 = 0
x = −4/3

Horizontal asymptote as x approaches ±∞
lim[x→±∞] f(x) = lim[x→±∞] (2x−5)/(3x+4) = 2/3

Vertical asymptote: x = −4/3
Horizontal asymptote: y = 2/3

f(x) = (2x - 5) / (3x + 4)

y-int.: x = 0:

y = [2(0) - 5] / [3(0) + 4]
y = - 5/4

y-int. (0, 0 5/4)
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f(x) is a fraction and, as such, the denominator cannot equal zero, lest division by zero occurs, which is undefined, so

3x + 4 ≠ 0
3x ≠ - 4
x ≠ - 4/3

Vertical Asymptote: x = - 4/3
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If the largest exponent in the numerator is equal to the largest exponent in the denominator, the horizontal asymptote is the coefficient of the variable with the largest exponent in the numerator divided by the coefficient of the variable with the largest exponent in the denominator.

Horizontal Asymptote: y = 2/3
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solving I'ts y - intercept

...........2x - 5
f(x) = --------------
...........3x + 4

when x = 0

.......2(0) - 5
y = -------------- = - 5/4
.......3(0) + 4

y - intercept is (0, - 5/4)


solving the horizontal asymptote

...........2x - 5
f(x) = -------------- = 2/3 < horizontal asymptote
...........3x + 4


for vertical asymptote
............2x - 5
f(x) = ----------------
...........3x + 4

solving for x from the denominator

3x + 4 = 0
3x = - 4
x = - 4/3

so

................2(-4/3) - 5..........-23/3
f(- 4/3) = -------------------- = ------------ <--vertical asymptote..//
................3(-4/3) + 4...........0

1.
(2x-5)/(3x+4)
=-5/4
(0, -5/4)

2.