Question

How do I graph these two problems?

Answer 1

Answer:

• See below and the attached graphs.

Explanation:

Question 3.a. Graph y = 3 - 7x and its inverse

1. y = 3 - 7x (the red line on the graph)

The relation y = 3 - 7x is a linear function (whose graph is a line).

When you compare that function with the general slope-intercept form of the line, y = mx + b, you get the slope (m) and the y-intercept (b).

Thus, the slope of the function is - 7, and the y - intercept is 3. You can use that information to graph it.

Also, you can graph a line by joining any two points. You can build a table of the points and choose any pair.

This is a table for the function y = 3 - 7x

   x       y

   0      3 - 7(0) = 3

   1       3 - 7(3) = 3 - 21 = - 18

  2       3 - 7(4) = 3 - 28 = - 25

Thus, you can use the points (0,3), (1, - 18) and (2, -25) to graph the function.

2. Inverse of y = 3 - 7x (the blue line on the graph)

To find the inverse relation you switch y and x and then solve for the new y:

• x = 3 - 7y

• 7y = 3 - x

• y = (3/7) - x/7

That is also a linear function. So, its slope is -1/7 and its y-intercept is 3/7.

You can build a table to graph the line that represents the function:

   x       y

   0      3/7 - 0/7 = 3/7

   1       3/7 - 1/7 = 2/7

   2      3/7 - 2/7 = 1/7

   3       3/7 - 3/7 = 0

Hence, you can use the points (0, 3/7), (1, 2/7), (2, 1/7), (3, 0) to graph that function.

Question 3.b. Graph y = 4x² - 2 and its inverse

3.  y = 4x² - 2 (the green vertical parabola on the graph)

This is a quadratic function whose general form is y = ax² + bx + c. From that you know:

• The graph is a parabola
• a = 4, b = 0, c = -2
• The graph open upwards because the coefficient of the quadratic term, i.e. a,  is positive (4)
• The y-intercept is: y = 4(0)² - 2 = - 2 ⇒ (0, -2)
• The vertex is a minimum point and it is at x = -b/(2a) = 0 ⇒ (0, -2)
• The x-intercepts are at y = 0:

         0 = 4x² - 2 ⇒ 4x² = 2 ⇒ x² = 2/4 = 1/2 ⇒ x = ± $\sqrt{2}/2$   ⇒

        $(\sqrt{2}/2,0)$   and  $(-\sqrt{2}/2,0)$

• There you have three points that you can use to graph the function:

4. Inverse of y = 4x² - 2 (the purple horizontal parabola on the graph)

To find the inverse relation you switch y and x and then solve for the new y:

• x = 4y² - 2

• 4y² = x + 2

• y² = (x + 2)/4

• $y=\pm \sqrt{x+2}/2$

That relation is not a function because some values of x (most of them) have two different outputs.

To graph it you might divide the function in two branches and build a table for each branch.

A. $y=\sqrt{x+2}/2$

            x            y

          - 2           0

            2            1

            7           3/2

           14            2

B. $y=-\sqrt{x+2}/2$

            x            y

          - 2           0

            2           - 1

            7          - 3/2

           14           - 2

The domain of both branches is x - 2 ≥ 0 ⇒ x ≥ - 2, because the radicand (the amount inside the square root) cannot be negative.