Questions: Question 9 0 / 1 pt (3) The graph of (y=frac-7 x-53 x+10) has a horizontal asymptote with equation (square) (enter the equation of the horizontal asymptote)

$Question 9 0 / 1 pt (3) The graph of (y=frac-7 x-53 x+10) has a horizontal asymptote with equation (square) (enter the equation of the horizontal asymptote)$

Transcript text: Question 9 $0 / 1$ pt (3) The graph of $y=\frac{-7 x-5}{3 x+10}$ has a horizontal asymptote with equation $\square$ (enter the equation of the horizontal asymptote) Question Help: Video Written Example Submit Question

Solution

Solution Steps

Step 1: Identify the degrees of the numerator and denominator

The given function is $ y = \frac{-7x - 5}{3x + 10} $. The degree of the numerator $-7x - 5$ is 1, and the degree of the denominator $3x + 10$ is also 1.

Step 2: Compare the degrees of the numerator and denominator

Since the degrees of the numerator and denominator are equal, the horizontal asymptote is determined by the ratio of the leading coefficients.

Step 3: Calculate the horizontal asymptote

The leading coefficient of the numerator is $-7$, and the leading coefficient of the denominator is $3$. Therefore, the horizontal asymptote is: \[ y = \frac{-7}{3} \]

Final Answer

$\boxed{y = \frac{-7}{3}}$

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