Can u guys help me quick plz I will make brainliest (I need help with 32)plzzzz

Please Help me Check the image below!!!!!

WITCHER [35]

Answer:

First question:

The graph of $y=\frac{3-2x}{2-3x}$ has a vertical asymptote at x = $\frac{2}{3}$ and a horizontal asymptote at y = $\frac{2}{3}$

Second question:

The graph of equation $y=\frac{1-3x}{2+x}$ has a horizontal asymptote at y = -3 ⇒ C

Step-by-step explanation:

The vertical asymptotes will occur at the values of x for which make the denominator is equal to zero

The horizontal asymptotes will occur if:

Both polynomials are the same degree, divide the coefficients of the highest degree terms
The polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote

First question:

∵ $y=\frac{3-2x}{2-3x}$

- To find the vertical asymptote equate the denominator by 0

to find the value of x

∵ The denominator is 2 - 3x

∴ 2 - 3x = 0

- Add 3x to both sides

∴ 2 = 3x

- Divide both sides by 3

∴ $\frac{2}{3}$ = x

∴ The graph has a vertical asymptote at x = $\frac{2}{3}$

To find the horizontal asymptote look at the highest degree of x in both numerator and denominator

∵ The denominator and the numerator has the same degree of x

- Divide the coefficient of x of the numerator and denominator

∵ The coefficient of x in the numerator is -2

∵ The coefficient of x in the denominator is -3

∵ -2 ÷ -3 = $\frac{2}{3}$

∴ The graph has a horizontal asymptote at y = $\frac{2}{3}$

The graph of $y=\frac{3-2x}{2-3x}$ has a vertical asymptote at x = $\frac{2}{3}$ and a horizontal asymptote at y = $\frac{2}{3}$

Second question:

The graph has a horizontal asymptote at y = -3

means the numerator and the denominator has same highest degree and the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator equal to -3