1. State the points before you start working. For example: "My points are \( (10,1) \) and (4,10). 2. Compute the slope, \( m \), using the slope formula. If you compute a decimal, make sure to convert it to afraction, so it is an exact answer. 3. Use the point-slope form of a line to create the equation of the line going through the two points. If you have a special case of a line that is not a function, then just write the equation of the line and explain why you can't use the point-slope formula. 4. If possible, put the line in slope-intercept form and identify the y-intercept value as an ordered pair. Again, for those with special cases, glve the equation of the line and the \( x \)-intercept. 5. Use Desmos or Graphmatica to graph your Unc. Adjust the axce so the y-Intercept is shown and labeled. If your line does not have a y-intercept, still graph it, show and tabel the x-Intercept, and explain why, in terms of graphing, it is not a function.

**1. State the points:** My points are \( (10,1) \) and \( (4,10) \). **2. Compute the slope \( m \):** The slope formula is \[ m = \frac{y_2 - y_1}{x_2 - x_1}. \] Substitute the points into the formula: \[ m = \frac{10 - 1}{4 - 10} = \frac{9}{-6} = -\frac{3}{2}. \] **3. Use the point-slope form to create the equation of the line:** The point-slope form of a line is \[ y - y_1 = m(x - x_1). \] Using the point \( (10,1) \) and the slope \( -\frac{3}{2} \): \[ y - 1 = -\frac{3}{2}(x - 10). \] **4. Put the line in slope-intercept form and identify the y-intercept:** First, distribute the slope: \[ y - 1 = -\frac{3}{2}x + \frac{3}{2} \times 10. \] Simplify the constant term: \[ \frac{3}{2} \times 10 = 15. \] So the equation becomes: \[ y - 1 = -\frac{3}{2}x + 15. \] Add 1 to both sides: \[ y = -\frac{3}{2}x + 16. \] The slope-intercept form is \[ y = -\frac{3}{2}x + 16. \] The y-intercept is when \( x = 0 \); thus, the y-intercept is \[ (0,16). \] **5. Graphing using Desmos or Graphmatica:** Plot the line \( y = -\frac{3}{2}x + 16 \) on a graphing tool (such as Desmos). Ensure that the y-intercept \( (0,16) \) is clearly shown and labeled. You will see that the line crosses the y-axis at \( (0,16) \). The graph verifies that the line is a function because for every value of \( x \) there is exactly one value of \( y \).