Graph of -1/2x+3 and ln(x-1)on the same axis

To graph $y = -\frac{1}{2}x+3$ and $y = \ln(x-1)$ on the same axis, here are the key features for each function:

1. Graph of $y = -\frac{1}{2}x+3$ (Linear Function)

• Slope ($m$): $-\frac{1}{2}$ (The line goes down 1 unit for every 2 units to the right).
• Y-intercept: When $x=0$, $y = -\frac{1}{2}(0)+3 = 3$. So, the y-intercept is $(0,3)$.
• X-intercept: When $y=0$, $0 = -\frac{1}{2}x+3 \implies \frac{1}{2}x = 3 \implies x = 6$. So, the x-intercept is $(6,0)$.

To sketch this line, plot the points $(0,3)$ and $(6,0)$ and draw a straight line through them.

2. Graph of $y = \ln(x-1)$ (Logarithmic Function)

• Domain: The argument of the natural logarithm must be positive, so $x-1 > 0 \implies x > 1$. The graph exists only for $x$ values greater than 1.
• Vertical Asymptote: The line $x=1$ is a vertical asymptote. As $x$ approaches 1 from the right, $y$ approaches $-\infty$.
• X-intercept: When $y=0$, $0 = \ln(x-1)$. This means $x-1 = e^0 = 1$. So, $x-1=1 \implies x=2$. The x-intercept is $(2,0)$.
• Other points:
  • When $x=e+1 \approx 3.72$, $y = \ln((e+1)-1) = \ln(e) = 1$. So, plot $(e+1, 1)$ or approximately $(3.72, 1)$.
  • When $x=e^2+1 \approx 8.39$, $y = \ln((e^2+1)-1) = \ln(e^2) = 2$. So, plot $(e^2+1, 2)$ or approximately $(8.39, 2)$.

To sketch this curve, draw a dashed vertical line at $x=1$ for the asymptote. Plot the x-intercept $(2,0)$ and the other points. Draw a smooth curve that approaches the asymptote $x=1$ downwards and increases slowly as $x$ increases.

Summary for Sketching:

1. Draw your x and y axes.
2. For $y = -\frac{1}{2}x+3$: Plot $(0,3)$ and $(6,0)$. Draw a straight line connecting them.
3. For $y = \ln(x-1)$:
   • Draw a vertical dashed line at $x=1$.
   • Plot $(2,0)$.
   • Plot $(3.72, 1)$ and $(8.39, 2)$.
   • Draw a curve that starts near the asymptote $x=1$ (going downwards), passes through $(2,0)$, and then curves upwards through the other points.