# Solutions to Algebra 2 Problems

__Complex Numbers____Quadratic Equations____Functions____Polynomials__-

Solution to Problem 5-1__Rational Expressions, Equations, Inequalities and Functions__

Set all terms to the same denominator

\( \dfrac{x^2+3x-5}{(x-1)(x+2)} - \dfrac{2}{x+2} - 1 = \dfrac{x^2+3x-5}{(x-1)(x+2)} - \dfrac{2(x-1)}{(x+2)(x-1)} - \dfrac{(x-1)(x+2)}{(x-1)(x+2)} \)

Add and simplify

\( = \dfrac{x^2+3x-5-2(x-1) - (x-1)(x+2)}{(x-1)(x+2)} = \dfrac{-1}{(x-1)(x+2)} \)

Solution to Problem 5-2

We start by stating that x = 1 and x = -2 cannot be solutions because these values make the denominator zero.

Multiply all terms of the equation by \( (x-1)(x+2) \) in order to eliminate the denominator.

\( \dfrac{- x^2+5}{x-1} (x-1)(x+2) = \dfrac{x-2}{x+2} (x-1)(x+2) - 4 (x-1)(x+2) \).

Simplify

\( (- x^2+5)(x+2) = (x-2) (x-1) - 4 (x-1)(x+2) \).

Write the equation with zero on the right side

\( (- x^2+5)(x+2) - (x-2)(x-1) + 4 (x-1)(x+2) = 0 \)

Expand and group the left side

\( -x^3+x^2+12x = 0 \)

Factor the left side

\( -x(x+3)(x-4) = 0 \)

Solve

Solution set:\( \{-3,0,4 \} \)

Solution to Problem 5-3

We start by stating that x = 1 and x = - 1 cannot be included in the solution set because these values make the denominator zero.

Rewrite the inequality with right hand side equal to zero.

\( \dfrac{1}{x-1}+\dfrac{1}{x+1} - \dfrac{3}{x^2-1} \ge 0\)

Set all terms to the same denominator

\( \dfrac{(x+1)}{(x-1)(x+1)}+\dfrac{(x-1)}{(x+1)(x-1)} - \dfrac{3}{(x-1)(x+1)} \ge 0\)

Group the terms on the left side

\( \dfrac{2x-3}{(x-1)(x+1)} \ge 0\)

The zeros of the numerator and denominator of the left side are

\( -1 , 1 , 3/2 \)

and they split the number line into 4 intervals

\( (-\infty , - 1) , (-1 , 1 ) , (1 , 3/2] \text{ and } [3/2 , +\infty) \)

Use test values or table of signs to determine the solutions set of the inequality which is the interval

\( (-1 , 1) \cup [3/2 , +\infty) \)

Solution to Problem 5-4

Group the terms in the right side to rewrite the given function as the ratio of two polynomials.

\( y = \dfrac{3x^2}{5 x^2 - 2 x - 7} + 2 = \dfrac{3x^2}{5 x^2 - 2 x - 7} + 2 \dfrac{5 x^2 - 2 x - 7}{5 x^2 - 2 x - 7} = \dfrac{13 x^2 - 4 x - 14}{5 x^2 - 2 x - 7} \)

The degree of the numerator and that of the denominator are equatl, therefore the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator and is given by

\( y = 13/5 \)

The vertical asymptotes are given by the zeros of denominator

\( 5 x^2 - 2 x - 7 = 0 \)

Solve the above equation to obtain the vertical asymptotes

\( x = - 1 \) and \( x = 7 / 5 \)

Solution to Problem 5-5

The rational function in part c) has the degree of the numerator is 3 and that of the numerator is 2 and therefore has an oblique asymptote. The oblique asymptote is the quotient of the division of the numeartor by the denominator of the given rational function.

\( y = -\dfrac{x^3 + 2x ^ 2 -1}{x^2- 2} = -(x + 2) - \dfrac{2x+3}{x^2 - 2} \)

The oblique asymptote has the equation

\( y = - (x + 2) \)

The point of intersection is found by first solving for x the equation

\( -\dfrac{x^3 + 2x ^ 2 -1}{x^2- 2} = - (x + 2) \)

The above equation has one solution given by

\( x = - 3/2 \)

We now calculate y using by substituting x by -3 / 2 in the equation of the function or the oblique asymptote

\( y = - ((-3 / 2) + 2) = - 1 / 2 \)

The point of intersection of the graph of the function in part c) and its oblique asymptote is at

\( (-3 / 2 , -1 / 2) \)

Solution to Problem 5-6

The given function could be rewritten as

\( f(x) = \dfrac{2x-2}{x-1} = \dfrac{2(x-1)}{x-1} = 2\) for \(x \ne 1\)

Assuming that the given graph has a hole at x = 1, the graph of the given function could be graph d) (red).

__Trigonometry and Trigonometric Functions____Logarithmic and Exponential Functions__

Solution to Problem 1-1

The conjuage of z is given by

z* = 2 + 3 i

Hence

z z* = (2 - 3i)(2 + 3i) = 4 + 9 = 13

Solution to Problem 1-2

Multiply numerator and denominator by the conjugate of the denominator 2 + i

\( \dfrac{1-i}{2-i} = \dfrac{(1-i)(2 + i)}{(2-i)(2 + i)}\)

Simplify

= 3 / 5 - (1 / 5) i

Solution to Problem 2-1

Expand and write the equation in standard form

\( x^2 + 3x + 5 = 0\)

Find the discriminant Δ.

\( \Delta = b^2 - 4 a c = (3)^2 - 4(1)(5) = -11\)

The discriminat is negative and therefore the equation has two complex solutions given by the quadratic formulas.

\( x_1 = \dfrac{-b+\sqrt\Delta}{2 a} = \dfrac{-3+i\sqrt{11}}{2} = -\dfrac{3}{2} + \dfrac{\sqrt{11}}{2} i\)

\( x_1 = \dfrac{-b+\sqrt\Delta}{2 a} = \dfrac{-3-i\sqrt{11}}{2} = -\dfrac{3}{2} - \dfrac{\sqrt{11}}{2} i\)

Solution to Problem 2-2

Write the given quadratic equation in standard form and find the discriminant.

\( -2 x^2 + m x - 2 m =0\)

\( \Delta = b^2 - 4 a c = m^2 - 4(-2)(-2m) = m^2 - 16 m \)

For a qudratic equation to have two complex solutions, its discriminat must be negative. We therefore need top solve the inequatlity \( \Delta \lt 0\)

\( m^2 - 16 m \lt 0 \)

The above inequality has the solution set given by the interval

\( (0 , 16) \)

The values of m for which the given equation has two complex number are all the values in the interval \( (0 , 16) \).

Solution to Problem 3-1

To evaluate \( f(a-1)\) we substitute \( x \) by \( a - 1 \) in \( f(x) \). Hence

\( f(a-1) = - (a-1)^2 + 3((a-1) - 1) \)

Expand and simplify

\( f(a-1) = -a^2+5a-7 \).

Solution to Problem 3-2

The domain of function f is the set of all real numbers so that f(x) has real values. For a square root to be real, the radicand must be non negative. Hence the need to solve the inequality

\( x^2-16 \ge 0 \)

The solution set to the above inequality which is the domain of f is given by

\( (-\infty , -4] \cup [4 , +\infty) \)

Solution to Problem 3-3

One way to find the range of the given quadratic function is to rewrite it in vertex form by completing the square.

\(f(x) = - x^2 - 2x + 6 = - (x^2 + 2x) + 6 \\ = -((x + 1)^2 - 1) + 6 = - (x + 1)^2 + 7\)

The graph of the given quadratic function is a parabola opening downward and with vertex at the point (-1,7). Hence the range is given by the interval

\( (-\infty , 7] \)

Solution to Problem 3-4

\( (f_o g)(x) \) is the composition of two functions and is given by

\( (f_o g)(x) = f(g(x)) \)

Hence

\( (f_o g)(a - 1) = f(g(a-1)) = f((a-1)^2 + 2) \\
= \sqrt{(a-1)^2 + 2 - 2} = \sqrt{(a-1)^2}= |a-1|\)

Since \( a \lt 1 \) is equivalent to \( a - 1 \lt 0 \) , the final answer can be simplified to

\( (f_o g)(a - 1) = |a-1| = - (a - 1) = - a + 1 \)

Solution to Problem 3-5

Functions in parts a), b), c) are even functions and therefore are not one to one functions.

Function in part e) is not a one to one because it is a periodic.

Let us show algebraically that the functions in part d) and f) are one to one functions using the contrapositive to the definition of one-to-one functions which is: Function f is a one to one if:

\( f(a) = f(b) \implies a = b \)

For function j in part d); write the equation \( j(a) = j(b) \)

\( 1/a + 2 = 1/b + 2 \)

Solve for a

\( a = b \)

Hence j(x) in part d) is a one to one.

For function l in part f); write the equation \( l(a) = l(b) \\

\( ln(a-1) + 1 = ln(b-1)+1 \)

Solve for a

\( ln(a-1) = ln(b-1) \)

ln(x) is a one to one function, hence

\( a - 1 = b - 1 \)

\( a = b \)

Hence l(x) in part f) is a one to one.

Solution to Problem 3-6

Write the given function as an equation with y = f(x)

\( y = \dfrac{-x+2}{x-1} \)

We now need to solve the above equation for x. Cross multiply to obtain

\( y (x - 1) = (-x + 2) \)

Expand

\( yx - y = - x + 2 \)

Add x and y to both sides and simplify

\( y x + x = 2 + y \)

Factor x out on the left side

\( x(y + 1) = 2 + y \)

Solve for x

\( x = \dfrac{2 + y}{y + 1} \)

Interchange x and y.

\( y = \dfrac{2 + x}{x + 1} \)

The inverse of f is given by.

\( f^{-1}(x) = \dfrac{2 + x}{x + 1} \)

Solution to Problem 3-7

Function f is even if f(x) = f(-x) and odd if f(x) = - f(-x).

Let us calculate f(-x), g(-x) and h(-x) and compare them to f(x), g(x) and h(x) respectively

\( f(-x) = - (-x)^3 = x^3 = - f(x) \)

hence f is odd

\( g(-x) = |- x|+ 2 = |x| + 2 = g(x) \)

hence g is even

\( h(-x) = \ln( - x - 1) \)

h(-x) is neither equal to h(x) nor to - h(x) and therfore function is neither even nor odd.

Solution to Problem 3-8

Function f has a zero at x = - 2 therefore

\( f(-2) = 0 \)

hence \(2f(2x - 5) \) has a zero for

\( 2x - 5 = 0 \)

Solve for x

\( x = 5/2 \)

Solution to Problem 3-9

Checking the interval of definition of the different parts of each function, function g in part b) have intervals and formulas that correspond to the givern graph.

Solution to Problem 3-10

Definition of the average rate of change (ARC) of function f as x changes from \( x = a\) to \( x = a + h \) is given by:

ARC \( = \dfrac{f(a+h) - f(a)}{a+h - a} \)

Simplify

\( = \dfrac{\dfrac{1}{a+h} - \dfrac{1}{a}}{h} = \dfrac{1}{h}(\dfrac{1}{a+h} - \dfrac{1}{a}) = \dfrac{-1}{a(a+h)}\)

Solution to Problem 4-1

Use long division to rewrite the divion as

\( \dfrac{-x^4+2x^3-x^2+5}{x^2-2} = -x^2+2x-3+\frac{4x-1}{x^2-2} \)

Then identify the quotient Q and remainder R as

Q \( = -x^2+2x-3 \) , R = \( 4x - 1 \)

Solution to Problem 4-2

Use long division to rewrite the divion as

\( \dfrac{4 x^2+2x-3}{2 x + k} = Q(x) + \dfrac{R}{2x + k} \)

which may also be written as

\( 4 x^2+2x-3 = (2x + k)Q(x) + R \)

Substitute x by -k / 2 in the above to get

\( 4 (-k / 2)^2+2(-k / 2)-3 = (2(-k / 2) + k) Q(-k / 2) + R \)

Expand and simplify taking into account that the term \( (2(-k / 2) + k) Q(-k / 2) \) is equal to zero

\( k^2-k-3 = R \)

Substitute R by -1 (given) and solve for k.

\( k^2-k-2 = 0 \)

The above equation has two solutions

\( k = 2 \) and \( k = - 1\)

Solution to Problem 4-3

Since \( (x - 2) \) is a factor of \(p(x) \), then

\( p(x) = (x - 2) Q(x) \)

Q(x) is obtained by division

\( Q(x) = \dfrac{p(x)}{x - 2} = -2x^3-12x^2-22x-12 = -2 (x^3+6x^2+11x+6 )\)

We now need to factor \(x^3+6x^2+11x+6 \) using the rational root theorem

The factors of the constant term 6 are: \( 1,2,3,6 \) and the factor of the leading coefficient 1 is 1.

Possible solutions are given by \( \pm \) the ratios of the factors of 6 and the factors of the leadinig coeffifient 1.

\( \pm \dfrac{ 1,2,3,6}{1} \)

We can easily check that x = - 1 is a zero of \(x^3+6x^2+11x+6 \); hence

\( x^3+6x^2+11x+6 = (x+1)Q'(x) \)

Using division we get

\( Q'(x) = \dfrac{x^3+6x^2+11x+6}{x+1} = x^2 + 5x + 6 \)

\( x^2 + 5x + 6 \) may now be aesily factored as

\( Q'(x) = x^2 + 5x + 6 = (x+2)(x+3)\)

We now factor p(x) starting from the begining

\( p(x) = (x - 2) Q(x) = -2 (x - 2) (x^3+6x^2+11x+6 ) = -2 (x - 2) (x+1)Q'(x) = -2 (x - 2) (x+1) (x+2)(x+3) \)

Solution to Problem 4-4

Use the difference of two squares to factor

\( 16 x^4 - 81 = (4x^2 - 9)(4x^2 + 9)\)

The term \( 4x^2 - 9 \) is a difference of two squares but we need to write \( 4x^2 + 9 \) as a difference of two squares using the imaginary unit \(i\).

\( (4x^2 - 9)(4x^2 + 9) = (4x^2 - 9)(4x^2 - (3i)^2) \)

Factor again using the difference of two squares twice.

\( 16 x^4 - 81 = (4x^2 - 9)(4x^2 + 9) = (4x^2 - 9)(4x^2 - (3i)^2) = (2x - 3)(2x + 3)(2x - 3i)(2x + 3i) \)

Solution to Problem 4-5

Rewrite the equation with right term equal to zero.

\( (x - 3)(x^2 - 4) - (- x + 3)(x^2 + 2x) = 0 \)

Factor \( (x - 3) \) out

\( (x - 3)(x^2 - 4 + x^2 + 2x) = 0 \)

Group

\( (x - 3)(2 x^2 + 2x - 4) = 0 \)

Solve the two equations

\( x - 3 = 0 \) gives a solution \( x = 3 \)

\( 2 x^2 + 2x - 4 = 0 \) gives solutions \(x = 1\) and \( x = -2 \)

The solution set of the given equation is given by

{-2, 1, 3}

Solution to Problem 4-6

Rewrite the inequality with right term equal to zero.

\( (x + 2)(x^2-4x-5) - (-x - 2)(x+1)(x-3) \ge 0 \)

Factor \( (x + 2) \) out

\( (x + 2) ( (x^2-4x-5) + (x+1)(x-3) ) \ge 0 \)

Factor \( x^2-4x-5 \)

\( x^2-4x-5 = (x+1)(x-5) \)

Substitute back into the last inequality

\( (x + 2) ( (x+1)(x-5) + (x+1)(x-3) ) \ge 0 \)

Factor completely

\( (x + 2) (x+1)(2x - 8) \ge 0 \)

The right hand term has 3 zeros

- 2 , - 1 and 4

The 3 zeros split the number line into 4 intervals

\( (-\infty , -2] , [-2 , -1] , [-1 , 4] \text{ and } [4 , +\infty) \)

Use test values or table of signs to determine the solutions set of the inequality which is the interval

\( [-2 , -1] \cup [4, \infty) \)

Solution to Problem 4-7

The graph is that of a polynomial of odd degree with leading coefficient negative and therefore the function in part b) cannot be the correct answer.

The functions in parts a) c) and d) have the same zero at x = 1 of multiplicity 2.

The equation in part a) has a second zero at x = - 2 of multiplicity 5 but its y-intercept is equal to - 32 which is different from the y intercept -4 of the graph shown and therefore cannot be the correct answer.

The equations in part c) and d) has each a second zero at x = - 2 of multiplicity 3 but the y-intercept in part d) is equal to - 8 and therefore cannot be the correct answer.

The equations in part c) has a y-intercept is equal to - 4 and therefore is the correct answer.

Solution to Problem 4-8

The given function has one real zero at x = 1 and therefore its graph has one x intrcept only at x = 1. Hence graphs a) (green) and b) (blue) cannot be the correct answer.

Since k is negative graph d) (black) cannot be the answer and the only possible answer could be graph c) (red).

If graph c) is a possible solution with y intercept equal to 4, then .

\( f(0) = 4 \)

Simplify and solve the equation \( f(0) = 4 \) for k

\( k ((0) - 1)((0)^2 + 4) = 4 \)

\( k = -1 \)

Solution to Problem 6-1

To each rotation correspond an angle of 2π radians. For 1000 rotations corresponds

\( 1000 \times 2\pi = 2000 \pi \) radians rotated in one minute

There are 60 seconds in one minute. Hence in 1 second, the wheel rotates

\( 2000 \pi / 60 = 104.7 \) radians per second.

Solution to Problem 6-2

The absolute value of the angle is larger than 2π; hence we first rewrite the angle as a sum of a special angle and a multiple of 2π

\( sec(-11\pi/3) = sec(- 12\pi/3 + \pi/3) = sec(- 4\pi + \pi/3) \)

we then simplify

\( = sec(\pi/3) = 1 / cos(\pi/3) = 1 / (1/2) = 2\)

Solution to Problem 6-3

180 degrees correspond to π; hence

\( 1200 \text{ degrees } = \dfrac{1200 \times \pi}{180} = \dfrac{20}{3} \pi \)

Solution to Problem 6-4

180 degrees correspond to π; hence

\( \dfrac{-7\pi}{9} = \dfrac{\dfrac{-7\pi}{9} \times 180 }{\pi} = -140 \) degrees

Solution to Problem 6-5

We first start with the rangle of \( -2\sin(-0.5(x - \pi/5)) \) which is a sine function with amplitude 2.

\( -2 \le -2\sin(-0.5(x - \pi/5)) \le 2 \)

We next add - 6 to all terms of the double inequality above to obtain a new double inequality

\( -2 - 6 \le -2\sin(-0.5(x - \pi/5)) - 6 \le 2 - 6 \)

Simplify to obtain the range of the given fuction

\( - 8 \le -2\sin(-0.5(x - \pi/5)) - 6 \le - 4 \)

The range of the given function is given by the interval

\( [-8 , -4] \)

For a function of the form \( y = a \sin(bx + c) + d\) the period P is given by

\( P = 2\pi / |b| = 2\pi / |0.5| = 4 \pi \)

Solution to Problem 6-6

One way to graph the given function is by successive transformations

We start by graphing \( y = cos(2x) \) , graph shown below

We next graph \( y = cos(2x - \pi/4) \) by shifting the graph of \( y = cos(2x) \) to the right by \(pi/8 \); graph shown below.

We next graph \( y = - cos(2x - \pi/4) \) , by reflecting the graph of \( y = cos(2x - \pi/4) \) on the x axis; graph shown below.

We next graph \( y = - cos(2x - \pi/4) + 2 \) , by shifting the graph of \( y = - cos(2x - \pi/4) \) 2 units up; graph shown below.

The last graph is close to the graph b) given above in the question; hence the answer is b)

Solution to Problem 6-7

One period is the distance of one cycle which is from x = 0 to x = 4π; hence the period is

\( 4 \pi \)

We also know the formula of the perdiod P = 2π/|b|; hence

\( 4 \pi = 2\pi/|b| \)

Solve for b to obtain

b = 0.5 or b = - 0.5

Let us use b = 1; hence we start by writing the function to be found as

\( y = a \sin(0.5 x + c) + d \)

Let ymax be the maximum value of y and ymin be the minimum value of y. The amplitude |a| is given by

\( |a| = (ymax - ymin)/2 = (-1 - (-3))/2 = 1 \)

Two possible values for a 1 and -1; use a = 1 and the function to be found becomes

\( y = \sin(0.5 x + c) + d \)

The value of d is given by

\( d = ( ymax + ymin ) /2 = (-1 + (-3))/2 = -2 \)

The function to be found is given by

\( y = \sin(0.5 x + c) - 2 \)

The graph that we are given is that of the above function with no horizontal shifting; hence c = 0 and the equation of the graph is

\( y = \sin(0.5 x) - 2 \)

Note however that we may obtain an infinite number of solutions by adding k(2\pi) to the argument of sin as follows:

\( y = \sin(0.5 x + k(2\pi)) - 2 \) , where k is an integer.

Solution to Problem 6-8

We need to solve the given trigonometric equation and select the smallest positive solution. Rewrite the given equation as

\( \cos (2x - \pi/4) = - 1/2\)

Let \( t = 2x - \pi/4 \) and rewrite our equation in terms of t.

\( \cos (t) = - 1/2 \)

Solve for t to find an infinite number of solutions

\( t_1 = 2 \pi/3 + k(2\pi) \) and \( t_2 = 4 \pi/3 + k(2\pi) \) , where \( k \) is any integer

Substitute \( t \) by \( 2x - \pi/4\) and solve for x

\( 2x_1 - \pi/4 = 2 \pi/3 + k(2\pi) \) and \( 2x_2 - \pi/4 = 4 \pi/3 + k(2\pi) \)

\( x_1 = \pi/3 + \pi/8 + k \pi = 11\pi/24 + k \pi \) and \( x_2 = 2 \pi/3 + \pi/8 + k(\pi) = 19\pi/24 + k \pi \)

We now select the smallest positive solution using different value of the integer k. k = 0 gives the smallest solution

\( x_1 = \pi/3 + \pi/8 = 11\pi/24 \)

Solution to Problem 6-9

In the numerator, use the identity cot(x) = cos(x) / sin(x) and factor out cos(x) in the two terms on the right to rewrite the given expression as

\( \dfrac{\cot(x)\sin(x) + \cos(x) \sin^2(x)+\cos^3(x)}{\cos(x)} = \dfrac{\sin(x)\cos(x) / \sin(x) + \cos(x) ( \sin^2(x)+\cos^2(x))}{\cos(x)} \)

Simplify and use the identity \( \sin^2(x)+\cos^2(x) = 1\) to rewrite the expression as

\( = \dfrac{\cos(x) + \cos(x)}{\cos(x)} \)

Simplify further

\( = 2 \)

Solution to Problem 7-1

Rewrite the given expression in ratios of like terms

\( \dfrac{4x^2 y^8}{8 x^3 y^5} = ( \dfrac{4}{8})( \dfrac{x^2}{x^3} )( \dfrac{y^8}{y^5} ) \)

Use exponent rules to simplify

\( = \dfrac{1}{2} x^{2 - 3} y^{8-5} = \dfrac{1}{2} \dfrac{y^3}{x} \)

Solution to Problem 7-2

Rewrite \( 9 \) as \( 3^2 \)

\( \dfrac{3^{1/3} 9^{1/3}}{4^{1/2}} = \dfrac{3^{1/3} (3^2)^{1/3}}{4^{1/2}} \)

Use exponent rules to simplify

\( = \dfrac{3^{1/3} 3^{2/3}}{4^{1/2}} = \dfrac{3^{1/3+2/3}}{2} = \dfrac{3^{1}}{2} = \dfrac{3}{2} \)

Solution to Problem 7-3

The logarithm and exponential functions of the same base are inverse of each other; hence

\( 2x - 4 = b^c \)

Solution to Problem 7-4

Use change of base formula on the term \( \log_3(a^2) \) to rewrite the given expression as

\( \log_a(9) \cdot \log_3(a^2) = \log_a(9) \cdot \dfrac{\log_a(a^2)}{\log_a(3)} \)

Rewrite 9 as \( 3^2 \) and use the formulas \( \log_a(x^2) = 2 \log_a(x)\) and \( \log_a(a^n) = n \)to simplify

\( = \log_a(3^2) \cdot \dfrac{2}{\log_a(3)} = 2 \log_a(3) \cdot \dfrac{2}{\log_a(3)} = 4\)

Solution to Problem 7-5

Rewrite the given equation with one side e

\( \log(\dfrac{x+1}{x-1}) = \log(x + 1)^2 \)

The logarithmic function is a one to one function and therefore the above equation gives the equation

\( \dfrac{x+1}{x-1} = (x + 1)^2 \)

Use the cross product to rewrite the equation as

\( (x+1) = (x-1)(x + 1)^2 \)

and solve

\( (x - )(1 - (x-1)(x+1)) = 0 \)

\( (x - 1)(2 - x^2) = 0 \)

Solutions to the above equation are

\( x = 1 \) , \(x = \sqrt2 \) . and \( x = - \sqrt 2 \)

The argument to a logarithmic function must be positive, hence we now need to check the solution found above into the given equation. The only solution that checks the given equation is

\(x = \sqrt2 \)

Solution to Problem 7-6

Let \( z = e^x \) and rewrite the equation in term of z.

\( z^2 + z - 6 = 0 \).

Solve for z to obtain

\( z = 2 \) and \( z = - 3 \)

Substitute z by e^x and solve for x

\( e^x = 2\) gives \( x = ln(2) \)

\( e^x = - 3\) has no solution for x

The given equation has one solution

\( x = ln(2) \)

Solution to Problem 7-7

As x becomes very large, the term \( e^{x-1} \) approaches zero and the only term left is

\( y = 2(-2) = - 4 \).

which is the equation of the horizontal asymptote of the given function

Solution to Problem 7-8

As the argument of the log function \( 2x - 6 \) approaches zero from the right, f(x) approaches very small (- infinity) values; hence the vertical asymptote is found by solving the equation

\( 2x - 6 = 0 \).

Solution: \( x = 3 \)

which is the equation of the vertical asymptote of the given function.

Solution to Problem 7-9

Functions in part A) and part C) are given by

\( y = 2 - 0.5^{2x-1} \) and \( y = 2 - 0.5^{-2x+1} \) respectively

They both have the same horizontal asymptote y = 2.

Function of the form \( 0.5^{2x} \) is an exponential function with base 0.5 and therefore is a decreasing function. However \(- 0.5^{2x}\) is an increasing function because of the - sign. The graph in part d) is inceasing and has a horizontal asymptote y = 2. Hence

Function A) corresponds to graph d) and function C) corresponds to graph a)

Functions in part B) and part D) are given by

\( y = 0.5^{2x-1} \) and \( y = 0.5^{-2x+1} \) respectively

They both have the same horizontal asymptote y = 0.

\( 0.5^{2x - 1} \) is an exponential function with base 0.5 and therefore is a decreasing function. However \( 0.5^{- 2x} \) is an increasing function because of the - sign in the exponent. The graph in part c) is decreasing and has a horizontal asymptote y = 0. Hence

Function B) corresponds to graph c) and function D) corresponds to graph b)

Solution to Problem 7-10

Functions in part A) is given by

\( y = 2+ln(x-2) \) and has a vertical asymptote at \( x-2 = 0 \) which gives \( x = 2 \)

The graph in part d) has a vertical asymptote at x = 2 and therefore corresponds to function A).

Functions in part C) is given by

\( y = -ln(-x) \) and has a vertical asymptote at \( - x = 0 \) which gives \( x = 0 \)

The graph in part a) has a vertical asymptote at x = 0 and therefore corresponds to function C).

Functions in part B) and D) are given by

\( y =-log_2(x+1)-1 \) and \( y = -log_3(x+1)-1 \) respectively

They both have a vertical at \( x = -1 \) and they also have the same y intercept (0,-1).

We need to find points and check them on the graph.

At x = 1, the function in part B) gives y = - 2. The point(1,-2) is on the graph c) and therefore the function in B) corresponds to the graph in c) and the function in part D) corresponds to the graph in part b).

## More References and Links

Algebra ProblemsStep by Step Math Worksheets Solvers

Math Problems and Online Self Tests.

Basic Rules and Properties of Algebra.

More Intermediate and College Algebra Questions and Problems with Answers .