Hints for solving the problems - Apply 11.1

Problem No. 2 - Which of the points below satisfy the function y = 3x - 6

Substitute each ordered pair into the given function and see is the resulting statement is true.

           (2, 0)    0 = 3(2) - 6       0 = 6 - 6        0 = 0     true – satisfies

           (0, -6)   -6 = 3(0) - 6     -6 = 0 - 6      -6 = -6    true – satisfies

           (3, -3)   -3 = 3(3) - 6     -3 = 9 - 6       -3 = 3    false

           (-1, -3)  -3 = 3(-1) - 6   -3 = -3 - 6       -3 = -9  false

            (1, -3)  -3 = 3(1) - 6     -3 = 3 - 6        -3 = -3  true – satisfies

Problem No. 8 - Make a table of at lease three ordered pairs that satisfy the equation y = -x + 3. Then use your table to graph the function.

Try to use numbers for x (the independent variable) so that when substituting into the equation will produce whole numbers for y (the dependent variable)

                            x            y

                            0            3         y = -0 + 3      y = 3

                            3            0         y = -3 + 3      y = 0

                            1            2         y = -1 + 3      y = 2

When you plot these points on the graph {(0, 3), (3, 0), (1, 2)}, you will have dots at those respective ordered pairs. Oh by the way, did you notice that the point (3, 0) crossed the x axis, which is called an x-intercept, and the point (0, 3) crossed the y axis, which is called a y-intercept.

Problem No. 14 - For the function     g(f) = -3t² + t - 4

         a.   g(-2)    g(-2)  =  -3(-2²) + (-2) - 4  =  -12 - 2 - 4  =  -18

         b.   g(-1)    g(-1)  =  -3(-1²) + (-1) - 4  =    -3 - 1 - 4  =    -8

         c.   g(0)     g(0)    =  -3(0²)  + (0)  - 4  =    0 + 0 - 4  =    -4

         d  .  g(1)    g(1)    =   -3(1²) + (1) - 4   =    -3 + 1 - 4 =    -6

         e.    g(2)     g(2)   =   -3(2²) + (2) - 4   =  -12 + 2 - 4 =   -14

Problem No. 18 - For the function   

       a.    f(1/2)        the answer

       b.      f(0)                the answer

       c.       f(3)                 the answer

       d.       f(-1)       the answer

       e.        f(-3)   the answer

The last problem is undefined, because a 0 came up in the denominator and you cannot divide by 0. If 0, as in problem a, came up in the numerator, the answer is 0.

Problem No. 20 - Use the graph of the function y = -x² + 9, shown in the book to find the domain and range.

The domain of a function is all of the real numbers, x, for which the function is defined and produces a real number, y. The domain will be the numbers moving left and right of the origin.

The range of a function is all of the possible y-values you can get from the x-values. The range will be the numbers moving up and down from the origin.

As you notice by the graph in the book, it is a parabola opening downward.

Also notice that as you move left and right, the x values get larger and there is no maximum or minimum point.

    So the domain is all real numbers:      (-∞, ∞}     the answer

Notice the range, it has a maximum point 9 (notice in the equation there was a 9). So this means the range will be from 9 going down to infinity.

     So the range is all real numbers less than or equal to 9   [9, -∞)    the answer

This is the interval notation (that you need to know), which means all real numbers less than or equal to from nine to negative infinity. The [ means it includes the nine. If there was a ( then it wouldn’t include the 9. After or before the ∞ there is never a [ , because you can’t include infinity. You can never get to infinity.

Problem No. 24 - Use the graph of the function , given in the book, to find its domain and range. Write the domain and range in interval notation. We already did this in problem No. 20.

Notice going left to right, there aren’t any numbers graphed on the left of 0.

So that means the domain must be from 0 continuing to infinity.

      The domain is:     [0, ∞)          the answer

Also notice the range starts at -3 and moves up to infinity.

      The range is:     [-3, ∞)       the answer

Problem No. 30 - Circle the functions below that are linear.

A linear function has x shown to the first power. It can’t have any x raised to a power greater or less that one. It also can’t have an exponent that is a variable or a variable in the denominator of a fraction.

                   f(x) = x² - 1     a quadratic equation not linear

                   f(x) = 3x + 5    a linear equation

                   f(x) =           x can be any number not linear

                   f(x) = x/3 + 1  a linear equation

                   f(x) = 7 - 3/x   x could be 0 not linear

Problem No. 32 - The graph of the function y = (1/2)x is shown in the book.

The given equation is y = (1/2)x. This equation crosses the y-axis at the point (0, 0).

Notice line A is parallel to the given equation (same slope) and has been moved up two units and crosses the y-axis at (0, 2). So the equation for line a, is the same as the original, except the two has to be added.

       Equation for line A     y = (1/2)x + 2        the answer

Line B, parallel to the given equation (same slope) has been moved down 5 units and crosses the y-axis at (0, -5).

       Equation for line B      y = (1/2)x - 5         the answer

Problem No. 36 - Graph the line    y = 2x + 4

This can be graphed by using a set of ordered pairs, using the x- & y-intercept or using the slope (m) y-intercept (b) method.

For this problem I will use the x & y-intercept.

                                x            y

                                0            4         y = 2(0) + 4      y = 4      (0, 4)

                               -2            0         0 = 2x + 4       -4 = 2x     x = -2     (-2, 0)

On graph paper graph the two ordered pairs, (0, 4) & (-2, 0). And then draw the line between then and extend it to the left and right of the two points to the end of the graph.

Problem No. 38 - Graph the line     y = (-1/2)x + 3

This will be graph using the slope (m) y-intercept (b) method.

The equation must be put in the form:   y = mx + b

     It already is, so:      m = -1/2   and   b = 3

This means you start at the point (0, 3). Then move two units to the right and one unit down and you are on the point: (2, -4). Now draw the line between the points and extend it at each end to the end of the graph.

Problem No. 42 - Graph the function    

This is a graph showing absolute value. There will be no points below the x-axis. All the values for y will be positive, because of the absolute value.

Plot this with a chart of points.

                 x            y

                -3            2                 (-3, 2)

                -2            1                  (-2. 1)

                -1            0                    (-1, 0)

                 0             1                      (0, 1)

                 1             2                     (1, 2)

                 2             3                     (2, 3)

                 3             4                     (3, 4)

Now on your graph paper plot the above points and draw straight lines from where the v starts. It starts at the point (-1, 0). Draw a straight line to the left and to the right from this point up.

Problem No. 46 - Find the domain and range of the function

                y = 10 - (1/5)x.

This is a linear function, which shows a straight line in its graph. That line will go left and right to infinity and up and down to infinity, so the;

        domain is:   (-∞, ∞)   and the range is:   (-∞, ∞)

This is the interval way of showing all real numbers.

Problem No. 50 - Find the domain and range of the function 

This graph will give you a v starting at the point (0, -1) and opening down to the left and the right. A v shape opening down.

  This shows the domain to be:   (-∞, ∞)   and the range to be   [-1, ∞).

Problem No. 54 - The graph of the function    is shown in the book. Graph the function    on your graph paper. It might help if you copied the graph from the book and then put the new graph next to it.

The original v starts at (0, -1). All you do is move the whole graph down two units to the v point (-1, -2). Then you draw straight lines up to the left and right which will be parallel to the lines of the original graph.

When the term v is used ,it is Mathematically called the maximum or minimum point. If the graph opens up it is the minimum point and if the graph opens downward it is the maximum point.

Problem No. 58 - Circle the functions below that are quadratic.

Quadratic functions are equations that have a squared term.

                     y = x + 10                   linear function

                     y = -5x² + 3x - 7         quadratic function   (circle)

                     y = 3x² + 2                 quadratic function   (circle)

                     y = 7 - 2x - 10x²         quadratic function   (circle)

                     y = 4 - x                      linear function

Problem No. 62 - Determine whether each of the following parabolas opens up or down.

A parabola will open up if the number in front of the quadratic term (a) is positive.

A parabola will open down if the number in front of the quadratic term (a) is negative.

                   a.     y = -2x²                        opens downward

                   b.     y = (5/7)x² - 3x + 7      opens upward

                   c.     y = 4 - 30 - 5x²             opens downward

                   d.     y = 6x² - 9                    opens upward

                   e.     y = 2x - x²                    opens downward

Problem No. 64 - Use the graph given in the book of the function

                    y = 6 - x²

To find the domain and range. Write the domain and the range in interval notation.

After looking at the graph, notice the numbers moving left and to the right have no restriction, so they go to infinity.

           The domain is:     (-∞, ∞)

The graph starts at (6, 0) and moves down, so the range is:    (-∞, 6].

Problem No. 68 - The graph of the function y = (x + 3)² is shown on the graph in the book. Find the following:

Notice equation A is moved up one unit, so the equation is:   y = (x + 3)² + 1.

Notice equation B is moved down three units, so the equation is:  y = (x + 3)² - 3.

Problem No. 74 - Find the x- and y-intercepts of the function   y = 2x² - 18.

   Determine the x- and y- intercepts by the following.

                               x               y

                               0            -18     y = 2(0)² - 18     y = 0 - 18     y = -18     (0, -18)

                              ±3             0      0 = 2x² - 18      18 = 2x²     x² = 9    x = ±3    (3, 0), (-3,0)

       So the x-intercepts are:      (3, 0) & (-3, 0).

           nd the y-intercept is    : (0, -18)

Problem No. 78 - The graph of the parabolas   y = -(x – 2)² - 1   and

  y = (x – 2)² - 1 are shown in the book. Use these graphs to decide which of the statements below are true.

Look at the graphs and you can tell that the graphs start at the vertex of each graph and both go to the left and the right to infinity, so they have the same domain.

The graph shows that one has y values going downward and the other goes upward, so that don’t have the same range.

The graphs also start at the same point, so the have the same vertex.

The y-intercept of one graph is (0, 3) and the other (0, -5). This shows the y-intercepts are not the same.

Problem No. 80 - The graph of the parabola   y = -x² - 4x - 3   is shown in the book. If the graph is moved up 3 units, what is the vertex of the new parabola?

The vertex of the original parabola is:  (-2, 1). You can determine this by looking at the graph.

So to the whole graph will be moved up three units and that means the vertex will be moved up 3 units. Add 3 to the y coordinate of the original vertex and you get (-2, 4) the answer.

Problem No. 84 - If the graph of the parabola   y = -x² + 6x - 5   is moved down 2 units, what is the equation of the new parabola?

If the parabola is moved upward you add to the constant (the last number) and if the parabola is moved down you subtract from the constant.

In this case the parabola is moved down 2 units, so you subtract 2 from -5 and you get an answer of:    y = -x² + 6x - 7