I see but the point l didn't seen show all point on the graph according to your number For example X/ y starting from 1up to 7 and point of it

You're right, sthamntungwa! Let's list out more specific points for plotting to make your graph even clearer.

Here are the calculations and coordinates for plotting $f(x)$, $f^{-1}(x)$, and the line $y=x$.

1. Line of Reflection: $y=x$

This line passes through points where the x-coordinate equals the y-coordinate.

• $\boxed{\mathbf{(0,0)}}$
• $\boxed{\mathbf{(1,1)}}$
• $\boxed{\mathbf{(2,2)}}$
• $\boxed{\mathbf{(3,3)}}$
• $\boxed{\mathbf{(4,4)}}$
• $\boxed{\mathbf{(5,5)}}$
• $\boxed{\mathbf{(6,6)}}$
• $\boxed{\mathbf{(7,7)}}$

2. Graph of $f(x) = 2x - 5$

The domain for $f(x)$ is $\{x: 0 < x \le 6, x \in \mathbb{R}\}$.

• At $x=0$: $f(0) = 2(0) - 5 = -5$. Point: $\boxed{\mathbf{(0, -5)}}$ (This is an open circle as $x > 0$).
• At $x=1$: $f(1) = 2(1) - 5 = -3$. Point: $\boxed{\mathbf{(1, -3)}}$
• At $x=2$: $f(2) = 2(2) - 5 = -1$. Point: $\boxed{\mathbf{(2, -1)}}$
• At $x=3$: $f(3) = 2(3) - 5 = 1$. Point: $\boxed{\mathbf{(3, 1)}}$
• At $x=4$: $f(4) = 2(4) - 5 = 3$. Point: $\boxed{\mathbf{(4, 3)}}$
• At $x=5$: $f(5) = 2(5) - 5 = 5$. Point: $\boxed{\mathbf{(5, 5)}}$
• At $x=6$: $f(6) = 2(6) - 5 = 7$. Point: $\boxed{\mathbf{(6, 7)}}$ (This is a closed circle as $x \le 6$).

3. Graph of $f^{-1}(x) = \frac{1}{2}x + \frac{5}{2}$

The domain for $f^{-1}(x)$ is $\{x: -5 < x \le 7, x \in \mathbb{R}\}$.

• At $x=-5$: $f^{-1}(-5) = \frac{1}{2}(-5) + \frac{5}{2} = -\frac{5}{2} + \frac{5}{2} = 0$. Point: $\boxed{\mathbf{(-5, 0)}}$ (This is an open circle as $x > -5$).
• At $x=-4$: $f^{-1}(-4) = \frac{1}{2}(-4) + \frac{5}{2} = -2 + 2.5 = 0.5$. Point: $\boxed{\mathbf{(-4, 0.5)}}$
• At $x=-3$: $f^{-1}(-3) = \frac{1}{2}(-3) + \frac{5}{2} = -1.5 + 2.5 = 1$. Point: $\boxed{\mathbf{(-3, 1)}}$
• At $x=-2$: $f^{-1}(-2) = \frac{1}{2}(-2) + \frac{5}{2} = -1 + 2.5 = 1.5$. Point: $\boxed{\mathbf{(-2, 1.5)}}$
• At $x=-1$: $f^{-1}(-1) = \frac{1}{2}(-1) + \frac{5}{2} = -0.5 + 2.5 = 2$. Point: $\boxed{\mathbf{(-1, 2)}}$
• At $x=0$: $f^{-1}(0) = \frac{1}{2}(0) + \frac{5}{2} = 2.5$. Point: $\boxed{\mathbf{(0, 2.5)}}$
• At $x=1$: $f^{-1}(1) = \frac{1}{2}(1) + \frac{5}{2} = 0.5 + 2.5 = 3$. Point: $\boxed{\mathbf{(1, 3)}}$
• At $x=2$: $f^{-1}(2) = \frac{1}{2}(2) + \frac{5}{2} = 1 + 2.5 = 3.5$. Point: $\boxed{\mathbf{(2, 3.5)}}$
• At $x=3$: $f^{-1}(3) = \frac{1}{2}(3) + \frac{5}{2} = 1.5 + 2.5 = 4$. Point: $\boxed{\mathbf{(3, 4)}}$
• At $x=4$: $f^{-1}(4) = \frac{1}{2}(4) + \frac{5}{2} = 2 + 2.5 = 4.5$. Point: $\boxed{\mathbf{(4, 4.5)}}$
• At $x=5$: $f^{-1}(5) = \frac{1}{2}(5) + \frac{5}{2} = 2.5 + 2.5 = 5$. Point: $\boxed{\mathbf{(5, 5)}}$
• At $x=6$: $f^{-1}(6) = \frac{1}{2}(6) + \frac{5}{2} = 3 + 2.5 = 5.5$. Point: $\boxed{\mathbf{(6, 5.5)}}$
• At $x=7$: $f^{-1}(7) = \frac{1}{2}(7) + \frac{5}{2} = 3.5 + 2.5 = 6$. Point: $\boxed{\mathbf{(7, 6)}}$ (This is a closed circle as $x \le 7$).

Use these points to draw your graph accurately, remembering to use open circles for points that are not included in the domain/range and closed circles for points that are included.