Answer

Q1. Which of the graphs describe y as a function of x.

From the given graphs, those that describe y as a function of x i.e. graphs which are functions are graphs B and C. this is because a graph is said to be a function if a vertical line drawn parallel to the y axis intersects the curve only ones. Thus, graphs A and D do not describe functions because a vertical drawn on the graphs would intersect the curves, four times and two times, respectively suggesting that their x-values have more than one output. As a rule of thumb, functions have one output value (y value) for every one input (x value).

Q2. Consider the points (-5, -6) and (-1, 10).

1. Midpoint of the line with the given end points

Midpoint of a straight line with given endpoints is expressed as;

M = ,        Where M is the midpoint, and x and y the given points.

Thus, with (-5, -6) and (-1, 10) as the given endpoints midpoint of the line can be   computed as;

 M =           = (-6/2, 4/2)

Hence, midpoint of the line is (-3, 2).

1. Length of the radius of circle using (-3, 2) as the center and (-5, -6) and (-1, 10) as points on the circle.

Distance between (-3, 2) and either of the points (-5, -6) and (-1, 10) i.e. radius of the circle can be obtained the Pythagorean theorem. Because (-2, 2) is midway between the endpoints, point (-1, 10) can be used as the other point between which to find distance.

Thus, we have;

From the above figure, distance of the radius of the circle formed by the two points can be computed as;

Distance =            =

            = √ (4+64) = √ 68 = 8.25 Units.

Q3. From the following graph of y = f(x), state;

1. X-intercept

Being a cubic function, the above graph has three x-intercepts i.e. points on the graph where y-value is zero. The intercepts are;

(-1, 0), (1, 0) and (-5, 0).

1. Y-intercept

y-intercept of the above function i.e. point of the graph where the values of x is zero is;

(0, -3)

1. Domain of the graph

Domain of the above function is (-1, 6]. This is because it extends horizontally from -1 to 6. 

1. Range

Range of the above graph is [-3, 2) because it has a vertical extent of 2 to -3.

Q4. Given  ;

1. Find f (-2)

Substituting x with -2 in the function;

We have f (-2) =  = -3/100

            f (-2) =

1. State domain of ;

Setting the denominator to 0 and solving for x, we have;

(x – 8)² = 0; which expands to

x² – 16x + 64 = 0.

Solving using quadratic formula:  ;

The roots for the equation become;  = (18 ± 0)/2 = 8

Thus, the domain of ;  = all real numbers of x except 8

In notational form, the domain is (∞, 8) U (8, ∞).

1. Evaluate f (a + 7)

Substituting x in the function with (a + 7), we get;

f (a + 7) =  =  =

Q5. Domain of the quotient function given  and

            Replacing x in the function g(x) with -7;

            We have g(x) = g(-7) = |-7 + 7| = |0| = 0

            Hence, domain of the quotient function is (-∞, -7) U (-7, ∞).

            This means the domain of the quotient are all real values of x except -7.

Question 6

1. Income of $6,300 lies at 2500<x≤7500

Therefore, T(x)= T($6,300)=70+0.035(x-2,500)=70+0.035(6,300-2,500)=$203

1. Income of $63,000 lies at x>7,500

Therefore, T(x)=T(63,000)=245+0.050(63,000-7,500)= $3,020

 Q7. Let

1. X-intercepts of the graph

Given that the value of y is zero at the x-intercept i.e. the line y = 0;

We have, 2 – 2x² = 0; simplifying and solving the equation leads to;

2x² = 2; =which translates to x² = 1; and hence, x = ±1.

Thus, x-intercepts of the graph are (1, 0) and (-1, 0).

1. Y-intercept of the graph.

Given that x value is always zero at the y-intercept; i.e. the line x = 0, we equate 0 in the function;

Hence, we have y = 2 – (2*0)² = 2;

This means the y-intercept of the graph is (0, 2)

1. Table of sample points on the graph of the equation

Figure 1: Graph of the equation y=2-2x² drawn in desmos graphing calculator.

1. Symmetry of the graph: the graph is symmetrical with respect to the y axis at points (-1, 0) and (1, 0); (-6, 0) and (6, 0) among other points.
2. The graph is not symmetrical around the x-axis.
3. The graph is symmetrical with respect to the origin as it passes through points (-1, 0) and (1, 0).

Q8. Let and

1. Evaluate g – f for x = -3

f(x) = 4x² + 2x – 8;            therefore f(-3) = 4(-3)² + 2*(-3) – 8 = 22

g(x) = 1 – 2x;                      therefore g(-3) = 1- (2*-3) = 7

thus, (g – f)(-3) = 7 – 22 = -15.

1. Evaluate fg for x = -3 i.e find fg(-3)

f(x) = 4x² + 2x – 8;            This implies f(-3) = 4(-3)² + 2*(-3) – 8 = 22

                g(x) = 1 – 2x;                      This implies g(-3) = 1- (2*-3) = 7

                Hence fg(-3) = 22*7 = 154

1. Find the difference function (f – g) (x)

(f – g) (x) = (4x² + 2x – 8) – (1 – 2x)

                = 4x² + 2x +2x – 8 – 1

Thus, (f – g) (x) = 4x² + 4x – 9

Question 9

7. Cost C(x) =5,250+7.00x for x≥0

When x=300, C(x)=C(300)=5250+7.00(300)=$7,350

This implies that the total cost of producing 300 widgets is $7,350

1. Average cost ‘C(x)=Total cost/Number of units produced

for X=300, ‘C(x)= ‘C(300)=7,350/300=$245

This means that the average cost of producing each unit of widget is $245. However, the average cost does not represent the actual cost of production for the item.

1. Revenue Function=price-demand function*total output in units

For x units, R(x)=x*p(x)=x(45-0.02x) for 0≤x≤2,250

R(x)=45x-0.02x² for 0≤x≤2,250

1. Profit=Total Revenue-Total Cost

Profit function, P(x)=45x-0.02x² -5,250-7.00x for 0≤x≤2,250

P(x)=-0.02x²+38x-5,250

1. For 300 units, P(x)=-(0.02*300²)+(38*300)-5,250=$4,350

This implies that by producing and selling 300 widgets, the company gains a total of $4,350