MATH 1090 Algebra and Trigonometry
Problem Set 5 - Solutions

2002 Winter

Note

Sketch the graphs of the following quadratic functions.
Indicate the locations of the vertex and all axis intercepts on your sketch.
1. y = 8 - 2(x + 3)²
  Þ - 2(x - (- 3))² + 8
  Compare with
  y = a (x - h)² + k
  
  a < 0 Þ (parabola opens downward)
  Vertex at ( h, k ) = ( - 3, 8 )
  
  x = 0 Þ y = - 2(3)² + 8 = - 2(9) + 8 = - 18 + 8 = - 10
  y-intercept is - 10
  y = 0 Þ - 2(x + 3)² + 8 = 0 Þ 2(x + 3)² = 8
  Þ (x + 3)² = 4 Þ x + 3 = 2
  Þ x = - 3 2
  x-intercepts are - 5, - 1
  Sketch:
2. y = 3x² - 6x + 5
  a > 0 Þ (parabola opens upward)
  
  k = y (1) = 3 - 6 + 5 = 2
  Vertex at (1, 2)
  Þ no x-intercepts
  
  y-intercept: x = 0 Þ y = 5
  Sketch:
3. y = 9 - x²
  y = - 1 x² + 9
  
  a = - 1 Þ (parabola opens downward)
  Þ vertex = y-intercept
  x = 0 Þ y = 9
  y = 0 Þ 9 - x² = 0 Þ x² = 9 Þ x = 3
  x-intercepts are 3
  Sketch:

300 metres of fencing is available to fence off a rectangular playing field that must also be divided into two areas by a fence parallel to a boundary fence. Find the lengths of the fences that maximizes the total enclosed area and find that maximum area.
Let x, y be the lengths
of fences as shown.
(3 fences, each length x,
2 fences, each length y)

Total length of fencing :
300 = 3 x + 2 y Þ 2 y = 300 - 3 x
Þ y = 150 -
Area:
A(x) = x y = x (150 - ) = 150 x - ²
But A = 150 x - ² is a parabola with a = < 0 ,
vertex at x =
The largest value of A occurs at the vertex.
x = 50 Þ y = 150 (50) = 150 - 75 = 75
and A = 150 (50) (50)² = 7500 - 3750 = 3750

Sketch the graph of the function f (x) = -2 (x - 4)² (x² - 25)
Seek just the leading coefficient.

f (x) = -2 (x² - ...) (x² - ...)
= -2 x² * x² + ...
= -2 x⁴ + ...

a ₄ Þ end-behaviour is
x = 0 Þ y = f(0) = -2 (-4)² (-25)
= -2 (16)(-25)
= 800 ( y-intercept )
y = 0 Þ (x - 4)² (x² - 25) = 0
Þ (x - 4)²(x - 5)(x + 5) = 0

Þ x = -5, 4, 4, 5
x = 4 is a root of multiplicity 2

y = f (x) crosses the x-axis at x = -5 and x = 5
It touches the x-axis and turns back up at x = 4
Sketch :

Note:
This sketch is not to scale.
Graphical computer/calculator
packages will show that the
peak to the right of x = 4
is less than 3 units above the
x-axis - invisible on any
scale that shows the y-intercept

For the function f (x) = x³ - 13 x² + 144 ,

Find the roots of the equation f (x) = 0 .

Possible rational roots : (factors of a0)/(factors of a3)

{+-1,+-2,+-3,+-4,+-6,+-9,+-12,+-16,+-24,+-36,+-48,+-72,+-144}/
{+-1}

f (1) = 1 - 13 + 144

0
f (-1) = -1 - 13 + 144

0
f (2) = 8 - 52 + 144

0
f (-2) = -8 - 52 + 144

0
f (3) = 27 - 117 + 144

0
f (-3) = -27 - 117 + 144 = 0 <--

(x + 3) is a factor of f (x)

(x^3 - 13x^2 + 0x + 144)/(x + 3) = x^2 - 16x + 48

-16x² + 0x
-(-16x^2 - 48x)

48x + 144

0

x² - 16x + 48 = (x - 4) (x - 12)
therefore

f (x) = (x + 3) (x - 4) (x - 12)
and its roots are x = -3, 4, 12

ALTERNATIVE SOLUTION , using synthetic division

c	x³	x²	x¹	x⁰
	1	-13	0	144
		1	-12	-12

	1	-12	-12	132	remainder 0

	1	-13	0	144
		-1	14	-14

	1	-14	14	130	X

	1	-13	0	144
		2	-22	-44

	1	-11	-22	100	X

	1	-13	0	144
		-2	30	-60

	1	-15	30	84	X

	1	-13	0	144
		3	-30	-90

	1	-10	-30	54	X

	1	-13	0	144
		-3	48	-144

	1	-16	48	0	REMAINDER = 0

(x + 3) is a factor of f (x)

Let q (x) = 1x² -16x + 48
Sign changes are present for q (x) : + - +
q (-x) = x² + 16x + 48
+ + +

therefore

there are no roots of q (x) below x = -3.
(Descartes' rule of sign).

Solving q (x) = 0 reveals q (x) = (x - 4) (x - 12)
and the solution for f (x), as before.

Sketch the curve y = f (x) .
x-intercepts are the zeroes of f: -3, 4, 12
y-intercept: f (0) = 0 - 0 + 144 = 144
End-behaviour: a₃ = 1 > 0 Þ

Show that the function x⁵ - x³ - 1 has a real zero between 1 and 2.
Use the intermediate value theorem to find an approximation for that zero, correct to one decimal place.
f (x) = x⁵ - x³ - 1

f (1) = 1 - 1 - 1 = -1 < 0
f (2) = 32 - 8 - 1 > 0
, by the I.V.T., a real zero must exist in (1, 2)

f (1.5) = (1.5)⁵ - (1.5)³ - 1 = 3.21... > 0
zero is in (1, 1.5)

f (1.3) = (1.3)⁵ - (1.3)³ - 1 = 0.51... > 0
zero is in (1, 1.3)

f (1.2) = (1.2)⁵ - (1.2)³ - 1 = -0.23... < 0
zero is in (1.2, 1.3)

f (1.25) = (1.25)⁵ - (1.25)³ - 1 = +0.09... > 0
zero is in (1.20, 1.25)

Correct to one d.p., the zero is

x = 1.2

There is also an Excel spreadsheet file, available here,
to demonstrate this solution.

Follow the seven steps on page 327 of the course textbook to graph each of the following rational functions.

p (x) = x
q (x) = x + 1

Step 1: symmetry

f (-x) f (x) NOT EVEN
f (-x) -f (x) NOT ODD

Step 2: y-intercept
Þ (0, 0) is on the curve

Step 3: x-intercept

p (x) = 0 Þ x = 0 only.
Þ The only intercept on either axis is (0, 0).

Step 4: vertical asymptote

q (x) = 0 Þ x + 1 = 0 Þ x = - 1

Step 5: horizontal asymptote

p (x) and q (x) have the same degree (1).
horizontal asymptote is Þ y = 1

Step 6: plot between x-intercepts / vertical asymptotes.

x -2 -1 - 0 1

x + 1 -1 0 1 2

y 2 DNE -1 0

Step 7: sketch

p (x) = 4x
q (x) = x² - 1 = (x - 1) (x + 1)

1) f (-x) = -4x / ((-x)^2 - 1) = -4x / (x^2 - 1)

= - f (x)
Þ f (x) is ODD (point-symmetry in the origin).

2) f (0) = 0 / (0-1)

= 0 Þ y-intercept = 0

3) p (x) = 0 Þ 4x = 0 Þ x-intercept = 0 only.

4) q (x) = 0 Þ (x - 1) (x + 1) = 0
Þ Vertical asymptotes at x =

1.

5) p (x) is degree 1. q (x) is degree 2.
Þ horizontal asymptote is the x-axis. (y = 0)

6)

x	-2	-1	-	0		1	2

4x				0	2	4	8
x² - 1				-1	-	0	3

y				0	-	DNE

Note: we may use symmetry to sketch the other half ( x < 0 )
of the curve.

7)

p (x) = 16x⁴ - 1 = (4x² - 1) (4x² + 1)
= (2x - 1) (2x + 1) (4x ² + 1)
q (x) = x²+ 1

1) f (-x) = = f (x)
Þ f (x) is EVEN (symmetry in the y-axis).

2) f (0) = = -1 Þ y-intercept = -1

3) p (x) = 0 Þ (2x - 1) (2x + 1) (4x² + 1) = 0
Þ x-intercepts =

4) q (x) = 0 Þ x² + 1 = 0 : no real solutions
no vertical asymptotes

5) p (x) is degree 4. q (x) is degree 2.
Þ no horizontal asymptotes

6) Use symmetry to sketch x < 0

x 0 1

16x⁴ - 1 -1 - 0 15

x² + 1 1 2

y -1 - 0

7)