Question

the graph of ( f(x) = \frac{(9 - 4x)^2}{2x^2 - 3x + 12} ) has a horizontal asymptote at ( y = square )

Explanation:

Step1: Identify Degrees of Numerator and Denominator

For the function \( f(x)=\frac{(9 - 4x)^2}{2x^2-3x + 12} \), first, find the degree of the numerator and the denominator.

• Expand the numerator: \( (9 - 4x)^2=81-72x + 16x^2 \), so the degree of the numerator (highest power of \( x \)) is \( 2 \).
• The denominator is \( 2x^2-3x + 12 \), so the degree of the denominator is also \( 2 \).

Step2: Apply Horizontal Asymptote Rule for Equal Degrees

When the degree of the numerator (\( n \)) is equal to the degree of the denominator (\( m \)), the horizontal asymptote is given by the ratio of the leading coefficients.

• Leading coefficient of the numerator (coefficient of \( x^2 \)): For \( 16x^2-72x + 81 \), the leading coefficient is \( 16 \).
• Leading coefficient of the denominator (coefficient of \( x^2 \)): For \( 2x^2-3x + 12 \), the leading coefficient is \( 2 \).

Step3: Calculate the Ratio of Leading Coefficients

The horizontal asymptote \( y \) is the ratio of the leading coefficients: \( y=\frac{16}{2}=8 \).

Answer:

\( 8 \)