Practice Exam 2

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Problem 1 

The number of people y in town who will buy a Lake Tahoe sweatshirt if it is offered at x dollars can be modeled by the equation

        50x + y  =  2000

A.  Find the y-intercept and interpret this number in the context of sweatshirt purchases.

Solution

To find the x-intercept, set y equal to zero and solve for x:

    50x + 0  =  2000

    50x  =  2000

Now divide both sides by 50 to get

    x = 40

We can interpret this by saying that when the price reaches $40 no (0) Tahoe residents will buy the sweatshirt.

C.  Find the slope and interpret this number in the context of sweatshirt purchases.

Solution

We put the equation into y intercept form by subtracting 50x from both sides:

    y  =  -50x + 2000

Now, we can read off the slope as

    m  =  -50  =  -50/1

We can interpret this by saying that for every $1 increase in the price, there will be 50 fewer sweatshirts sold.  (as x increases by 1, y decreases by 50)

Problem 2

Find the domain and range of each relation.  Then determine whether it is a function.

A.  {(1,2), (2,4), (4,7), (4,8)}

Solution

The domain is the x-coordinates:  {1,2,4}

The range is the y-coordinates:  {2,4,7,8}

B.  x + 6y  = 12

Solution

This is a line that is neither vertical nor horizontal.  Hence the domain and the range are both all real numbers.

C.   

Solution

By reading the graph, we see that all x-values are obtained on the graph, hence the domain is all real numbers.

The range is the set of y-values that are achieved.  Notice that the graph does not go below y = -3, but goes upward without bound.  Hence the range is the set of real numbers from -3 to infinity.

Problem 3

A.  If f(x) = x² - 2x + 1, find f(-3).

Solution

To find f(-3) we just plug in -3 for x:

        f(-3) = (-3)² - 2(-3) + 1  =  9 + 6 + 1  =  16

B.  If the graph of g(x) is shown in problem 2C, find g(4).

g(4) is just the y-value when x is four.  You can go to the graph and see if you move to the right by 4, you must move downward 2.  That is

g(4) = -2

Problem 4

Sketch the graphs of the following lines and find the x and y intercepts is they exist.

A.  2x - 5y = 10

Solution

We find set x = 0 and find y:

    2(0) - 5y  =  10

        -5y  =  10        Divide both sides by -5

        y  =  -2

Next set y = 0 and find x

        2x - 5(0)  =  10

        2x  =  10

        x  =  5

This gives the table

The plot the points and connect with a line.  The x-intercept is 5 and the y-intercept is -2.  The graph is shown below.

B.  y = 4

Solution

This is just the horizontal line four units above the x-axis.  There is no x-intercept and the y-intercept is 4.

C.  y + 2/3 x  =  4

Solution

We can subtract the 2/3 x from both sides to get

        y  =   -2/3 x + 4

which is a line with slope -2/3 and y-intercept 4.  We can begin at the point (0,4), "rise" down 2 and "run" across 3.  To find the x-intercept, set y = 0 and solve.

        0  =  -2/3 x + 4

        0  =  -2x + 12

        2x  =  12

        x  =  6

The x-intercept is 6.

The graph is shown below.

D.  x = -3

Solution

This is the vertical line that crosses the x-axis at x = -3.  The x-intercept is -3 and there is no y-intercept.  The graph is shown below.

E.  The line that passes through the point (2,4) and has slope -3/2.

Solution

We just sketch the point (2,4) then "rise" down 3 and "run" across 2.  The graph is shown below.

To find the intercepts, first find the equations using the point-slope formula.

        (y - 4)  =  -3/2(x - 2)

        y - 4  =  -3/2 x + 3

        y  =  -3/2 x + 7

We see that the y-intercept is 7.  For the x-intercept, set y = 0 and solve.

        0   =  -3/2 x + 7

        0  =  -3x + 14

        3x  =  14

        x  =  14/3

So the x-intercept is at x = 14/3.

Problem 5

A.  Find the slope of the line that is parallel to the line that passes through the points (2,-3) and (5,6).

Solution

Parallel lines have the same slope, so we need only to find the slope through the points (2,-3) and (5,6).

The rise is:

        6 - (-3)  =  6 + 3  =  9

and the run is:

        5 - 2  =  3

Hence the slope is 

        m  =  rise/run  =  9/3  =  3

B.  Find the slope of the line that is perpendicular to the line that passes through the points (0,2) and (2,3).

Solution

First, we find the slope of the line through the two points.  

The rise is:

        3 - 2  =  1

and the run is:

        2 - 0  =  2

Hence the slope is 

        m  =  rise/run  =  1/2

Our line is perpendicular, so we take the reciprocal of the this slope and multiply by (-1) to get

        (-1) (2/1)  =  -2

Problem 6

Find the equation of the line in slope intercept form with the following properties.

A.  Passes through the points (2,3) and (4,-1).

Solution

First we find the slope.  The rise is

        -1 - 3  =  -4

and the run is 

        4 - 2  =  2

Hence the slope is

        m  =  rise/run  =  -4/2  =  -2

Now use the point slope formula with the point (2,3) and the slope -2 to get

        y - 3  = (-2)(x - 2)  =  -2x + 4        Add 3 to both sides

        y  =  -2x + 7

B.  Passes through the point (-4,2) and has slope -2/5.

Solution

We use the point slope formula to get

        y - 2  =  (-2/5)(x - (-4))  =((-2/5)(x + 4)  =  -2/5 x - 8/5  Add 2 to both sides

        y  =  -2/5 x - 8/5 + 2  =  -2/5 x - 8/5 + 10/5  =  -2/5 x + 2/5

or

        y  =  -2/5 x + 2/5

Problem 7

The graphs of several lines are shown below.  decide which of the statements are true and which are false and explain your reasoning.

A.  The slope of line II is greater than the slope of line I.

Solution

False.  The slope of line II is negative and the slope of line I is positive.  Negative numbers are always smaller than positive numbers.

B.  The y-intercept of lines I and II are equal.

Solution

True.  They are both equal to 2.

C.  The x-coordinate of the x-intercept of line I is greater than the x-coordinate of the x intercept of line III.

Solution

False.  Line I has negative x-intercept and line III has negative x-intercept.

D.  Lines I and II are perpendicular. 

Solution

False.  The slope of line I is 2/2  =  1 and the slope of line II is -2/1  =  -2.  Since their produce is 

        (1)(-2)  =  -2

which is not -1 they are not perpendicular.

Problem 8

Simplify the expressions below.

A.         p³y⁴ 
                    
           p²y

Solution

We use the division rule of exponents.  That means we subtract the exponents to get

     py³

B.  5¹ - (xyz)⁰ + 3⁰

Solution

Recall that any number to the 1 is itself and any number except 0 to the 0 power is 1.  This gives

        5 - 1 + 1  =  5

C .  (-2x²yz³)³

Solution

Use the power rule for exponents

      -8x⁶y³z⁹

Problem 9

Find

A.  (x³yz - 3z³ + 2yz² + x) + (2x³yz + 5z³ - 4x) - (2z³ - 5yz² + x)

Solution

We add the polynomials remembering to multiply the minus sign through.

      x³yz - 3z³ + 2yz² + x + 2x³yz + 5z³ - 4x - 2z³ + 5yz² - x

Now combine like terms

      3x³yz + 7yz² - 4x

B.  5a²b(2ab² - a³b + 3a⁷b⁵)

Solution

Multiply the monomial by the polynomial remember to add the exponents.

        10a³b³ - 5a⁵b² + 15a⁹b⁶

Problem 10 

A.  (3x - 4)(7x + 2)

We FOIL:

        F  =  21x²    O  =  6x    I  =  -28x    L  =  -8

Now add to get

        21x² + 6x - 28x - 8

Now combine like terms

        21x² - 22x - 8

B.  (x² + 2x)(3x³ - 4x + 5) 

Just distribute both of the first terms into each of the three last terms to get

        3x⁵ - 4x³ + 5x² + 6x⁴ - 8x² + 10x

Now combine like terms

        3x⁵ + 6x⁴ - 4x³ - 3x² + 10x

C.  (x - 2y)(x + 2y)

Solution

This is just a difference of squares with a = x and b = 2y.  We get

        x² - (2y)²

        =  x² - 4y²

D.  (4x + 3y)²

Solution

This is just a square of a sum with a = 4x and b = 3y.  We get

        (4x)² + (2)(4x)(3y) + (3y)²

        =  16x² + 24xy + 9y²

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