answers.mws

1. State, in your own words, the definition of the following terms:
a) slope of a line through two points (x, y) and (x, y)

If P1 = (x1, y1) and P2 = (x2, y2) then the slope m = (y2 - y1) / (x2 - x1)


b) function

A function is a rule that assigns to every number in its domain exactly one outcome number.

c) range and domain

Domain: All numbers for which a function is defined.

Range: All possible outcome numbers of a function.

d) linear function

A function with two variables, each raised to a power of at most 1.

e) monomial

Products of integer powers of variables and numbers

f) polynomial

Sums of monomials

2. Find the slope and y-intercept of the following lines, if possible, and graph them in the coordinate system provided:

2a)

f(x) := -3/5*x+12;

f(x) := -3/5*x+12

Has slope -3/5, y-intercept 12. The graph is:

> plot(-3/5*x+12, x=-3..3);

[Maple Plot]

2. b)

> solve(-5*y - 2*x = 7, y);

-2/5*x-7/5

Slope is -2/5, y intercept -7/5. The graph is

> plot(-2/5*x - 7/5, x = -5..5);

[Maple Plot]

2. c)

> solve(y + 5 = 7, y);

2

slope is zero, y intercept 2. The graph is:

> plot(2, x=-3..3);

[Maple Plot]

2. d)

> solve(3 = x - 1,x);

4

Vertical line (undefined slope) through x = 4.

3. Find the linear equation describing the following lines
a) line with slope 4 containing (-2, -4)

y + 4 = 4 (x + 2)

3. b) line through (3, -1) and (4, -2)

> m := (-2 - (-1)) / (4 - 3);

m := -1

> y +2 = m * (x - 4);

y+2 = m*(x-4)

>

c) line containing (-3, 2) and parallel to the line 2x - 5y = 8

> solve(2*x - 5*y = 8, y);

2/5*x-8/5

> m := 2/5;

m := 2/5

> y - 2 = m * (x + 3);

y-2 = 2/5*x+6/5

>

d) line containing (-3, 2) and perpendicular to the line 2x - 5y = 8

> m := -1 / (2/5);

m := -5/2

> y - 2 = m * (x + 3);

y-2 = -5/2*x-15/2

>

e) line through (1, 2) and perpendicular to the line through (-2, -2) and (3, 1)

> m := (1 - (-2)) / (3 - (-2));

m := 3/5

> m := -1 / (3/5);

m := -5/3

> y - 2 = m * (x - 1);

y-2 = -5/3*x+5/3

f) line through (-2, -2) and (3, -2)

m = 0, so it's a horizontal line passing through y = -2, so the equation is y = -2

g) line through (-2, -2) and (-2, 3)

m = 1/0 = undefined, so it's a vertical line passing through x = -2, so the equation is x = -2

4. For the following problems, evaluate the given function at the indicated position.

> f := x -> (x - 3) / (2*x - 5);

f := proc (x) options operator, arrow; (x-3)/(2*x-5...

> f(0);

0

> f(3);

9

> f(-1);

4/7

> f(a+h);

(a+h-3)/(2*a+2*h-5)

> g := t -> 5*t^2 + 4*t;

g := proc (t) options operator, arrow; 5*t^2+4*t en...

> g(0);

0

> g(-1);

1

> g(2*a);

20*a^2+8*a

> (g(a+h) - g(a-h)) / h;

(5*(a+h)^2+8*h-5*(a-h)^2)/h

> simplify(%);

20*a+8

5. Which of the following graphs represent a functions, which ones not? For each graph representing a function, determine the range and domain of that function.

First is a function. Domain is all numbers, range is all numbers (it's hard to see, but that's what it seems like).

Second is not a function.

Third is a function, domain is -1 to 1, range is 0 to 1

Fourth is a function, domain is -1 to 2, range is the numbers {1, 2}

6. Solve the following systems of linear equations.

> solve( {2*x-3*y=0, -4*x+3*y=-1}, {x,y});

{x = 1/2, y = 1/3}

> solve( {2*x+y=6, 3*x+4*y=4}, {x, y});

{x = 4, y = -2}

> solve( {9*x-6*y=2, x=4*y+5}, {x,y});

{x = -11/15, y = -43/30}

> solve( {0.2*x+0.3*y=1.7, 1/7*x+1/5*y=-29/35}, {x,y});

{x = -206., y = 143.}

> solve( {3*y-2*x=6, 8*x-12*y=-24}, {x,y});

{y = 2/3*x+2, x = x}

This means there are infinitely many solutions, or the lines are the same

> solve( {y=x+2, y-x=8}, {x,y});

This means there are no solutions, or the lines are parallel but not identical.

7. Solve the following inequalities and draw the solution set on a number line.

> solve(-4*y-3>=5, y);

RealRange(-infinity,-2)

> solve( -8*(2*x+3) + 6*(4-5*x) >= 2*(1-7*x)-4*(4+6*x), x);

RealRange(-infinity,7/4)

> solve(-15 < -4*x-5, x);solve(-4*x-5<1);

RealRange(-infinity,Open(5/2))

RealRange(Open(-3/2),infinity)

This means that the answer is (-3/2, 5/2).

> solve(-1/3 <= 1/6*x-1);solve(1/6*x-1<1/4);

RealRange(4,infinity)

RealRange(-infinity,Open(15/2))

This means that the answer is (15/2, 4)

> solve(3*x-2 < 7);solve(x-2 > 4);

RealRange(-infinity,Open(3))

RealRange(Open(6),infinity)

This means that the answer is (-infinity, 3) union (6, infinity)

> solve(abs(4*x-1) < 4.5);

RealRange(Open(-.8750000000),Open(1.375000000))

> solve(abs(-5*t-3) > 10);

RealRange(Open(7/5),infinity), RealRange(-infinity,...

> solve(abs(2.1*x-7.9) < -2);

This means that there is no answer (obviously, since an absolute value is never negative)

>

8. Find the domains of the following functions:

> f(x) := (x-2)/(3-x);

f(x) := (x-2)/(3-x)

Domain is all numbers except 3.

> g(x) := sqrt(7-2*x);

g(x) := sqrt(7-2*x)

> solve(7-2*x >= 0, x);

RealRange(-infinity,7/2)

That's the domain (-infinity, 7/2].

> h(x) := sqrt(2*x-4) / (x-4);

h(x) := (2*x-4)^(1/2)/(x-4)

> solve(2*x-4 >= 0, x);

RealRange(2,infinity)

Thus, the domain is all numbers bigger than or equal to 2, except 4 (because of the denominator).

9. Perform the following operations

> expand((4*x-1) * (4*x+1));

16*x^2-1

> expand((a+b-1)*(a+b+1));

a^2+2*a*b+b^2-1

> expand((a^2+a-1)*(a^2+a*b+b^2));

a^4+a^3*b+a^2*b^2+a^3+a^2*b+a*b^2-a^2-a*b-b^2

> expand((x-2*y)*(x^2+1)*(3-2*y^2));

3*x^3-2*x^3*y^2+3*x-2*x*y^2-6*y*x^2+4*y^3*x^2-6*y+4...

> factor(-2*x^3 + 4*x^2 - 2*x);

-2*x*(x-1)^2

> factor(12*a^4-21*a^3-9*a^2);

3*a^2*(4*a^2-7*a-3)

> factor( 5*x^2*(x-6) + 10*x*(6-x));

5*x*(x-2)*(x-6)

> f := x -> x^2;

f := proc (x) options operator, arrow; x^2 end proc...

10. If a given function f has a graph as indicated, please graph the modified function

> plot({f(x), f(x)-2, f(x+1), f(x-2)+1, -f(x), f(-x)}, x=-4..4);

[Maple Plot]

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