Question

sketching rational functions
( t(x) = \frac{2x - 3}{x - 2} )
find the following for the given rational function:
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
rough sketch of t(x)

Explanation:

Step1: Find x - intercepts

To find the x - intercepts, we set \(t(x)=0\). So we solve the equation \(\frac{2x - 3}{x - 2}=0\). A fraction is zero when its numerator is zero (and the denominator is not zero). So we set \(2x-3 = 0\). Solving for \(x\), we get \(2x=3\), so \(x=\frac{3}{2}\). We check that when \(x = \frac{3}{2}\), the denominator \(x - 2=\frac{3}{2}-2=\frac{3 - 4}{2}=-\frac{1}{2}
eq0\). So the x - intercept is at \(x=\frac{3}{2}\), and the point is \((\frac{3}{2},0)\).

Step2: Find y - intercepts

To find the y - intercepts, we set \(x = 0\) in the function \(t(x)\). So \(t(0)=\frac{2(0)-3}{0 - 2}=\frac{-3}{-2}=\frac{3}{2}\). So the y - intercept is at \(y=\frac{3}{2}\), and the point is \((0,\frac{3}{2})\).

Step3: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero). We set the denominator \(x - 2=0\), so \(x = 2\). We check the numerator at \(x = 2\): \(2(2)-3=4 - 3 = 1
eq0\). So the vertical asymptote is \(x = 2\).

Step4: Find horizontal asymptotes

For a rational function \(t(x)=\frac{ax + b}{cx + d}\) (where \(a,b,c,d\) are constants), if the degrees of the numerator and denominator are equal (both degree 1 in this case, since the numerator \(2x-3\) is degree 1 and the denominator \(x - 2\) is degree 1), the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is \(2\) and the leading coefficient of the denominator is \(1\). So the horizontal asymptote is \(y=\frac{2}{1}=2\).

Step5: Rough sketch

• The x - intercept is at \((\frac{3}{2},0)\) and the y - intercept is at \((0,\frac{3}{2})\).
• The vertical asymptote is \(x = 2\) (a vertical line).
• The horizontal asymptote is \(y = 2\) (a horizontal line).
• We can also analyze the behavior of the function around the vertical asymptote. For \(x>2\), as \(x\to2^{+}\), the numerator \(2x - 3\to2(2)-3 = 1\) and the denominator \(x - 2\to0^{+}\), so \(t(x)\to+\infty\). As \(x\to+\infty\), \(t(x)=\frac{2x-3}{x - 2}=\frac{2-\frac{3}{x}}{1-\frac{2}{x}}\to\frac{2}{1}=2\). For \(x<2\), as \(x\to2^{-}\), the denominator \(x - 2\to0^{-}\) and the numerator \(2x - 3\to1\), so \(t(x)\to-\infty\). As \(x\to-\infty\), \(t(x)=\frac{2x-3}{x - 2}=\frac{2-\frac{3}{x}}{1-\frac{2}{x}}\to2\). We can plot the intercepts, draw the asymptotes, and then sketch the two branches of the rational function: one to the left of \(x = 2\) (passing through \((0,\frac{3}{2})\) and approaching the asymptotes) and one to the right of \(x = 2\) (passing through \((\frac{3}{2},0)\) and approaching the asymptotes).

Answer:

• X - intercepts: \(\boldsymbol{(\frac{3}{2},0)}\)
• Y - intercepts: \(\boldsymbol{(0,\frac{3}{2})}\)
• Vertical asymptotes: \(\boldsymbol{x = 2}\)
• Horizontal asymptotes: \(\boldsymbol{y = 2}\)
• Rough sketch: Plot the intercepts \((\frac{3}{2},0)\) and \((0,\frac{3}{2})\), draw the vertical line \(x = 2\) and horizontal line \(y = 2\) as asymptotes. The graph has two branches: one left of \(x = 2\) (decreasing from \(y = 2\) (as \(x\to-\infty\)) to \(-\infty\) as \(x\to2^{-}\), passing through \((0,\frac{3}{2})\)) and one right of \(x = 2\) (increasing from \(+\infty\) as \(x\to2^{+}\) to \(y = 2\) as \(x\to+\infty\), passing through \((\frac{3}{2},0)\)).