Precalc JeopardyParametric Equations

Factoring/zeros Rational Functions

Conic Sections Polar graphs and complex

numbers

10 10 10 10 10

20 20 20 20 20

30 30 30 30 30

40 40 40 40 40

50 50 50 50 50

Find the parametric equations of the inverse of

2

5

2 7

x t

y t

22 7

5

x t

y t

To find the parametric equations of the inverse of any set of parametric equations just switch the x and y equations:

Give a set of parametric equations for a line through the points (2, 5) and (-3, 12).

The change in x-coordinates and y-coordinates of the points (2, 5) and (-3, 12) are: change in x: -5, change in y: 7

So then as t increases, these are the amounts that x and y increase by. We can start at either point, but let’s pick (2, 5).

5 2

7 5

x t

y t

Find the parametric equations of the line through the points: (3, -4) when t = 0 s, and (6, 2) when t = 2 s.

The change in x values is 3, but that happens over 2 second interval. So as t increases by 1 s, then x increases by 1.5. The change in y values is 6, but that happens over 2 second interval. So as t increases by 1 s, then y increases by 3. So the equations look like: 1.5 3

3 4

x t

y t

Jake throws a football in the air for his brother Justin to catch. It is thrown with the path of the football following the set of parametric equations:Jake throws it too short. Justin is 100 feet from Jake. How far in front of Justin did it hit the ground?

Solving for t when the y equation is zero, 0 = -16t2 + 46sin(31)t + 3,t = 1.598 sec. Then plugging this into the x equation to find where it hits the ground, x = 46cos(31)*1.598 63.011 ft so that means it was

36.989 ft in front of Justin.

2

46(cos31 )

16 46(sin31 ) 3

x t

y t t

Write the parametric equations for the graph of the hyperbola shown.

Knowing the center and the dilations…

4

2

-2

5

g x = -x-4 2

9+1+2

4 3tan

2 sec

x t

y t

Factor the following cubic:

f (x) = 3x3 – 5x2 – 2x

f (x) = 3x3 – 5x2 – 2x = x(3x2 – 5x – 2) = x(3x + 1)(x – 2)

Find all the zeros (real and non-real)of the cubic: f (x) = 2x3 +13x2 + 12x – 32

Then using the leftover quadratic:2x2 +5x – 8 …using the quadratic formula

Trying synthetic division we get:-4 2 13 12 -32 -8 -20 32 2 5 -8 0

25 5 4(2)( 8) 5 25 64 5 89

2(2) 4 4x

The zeros are: - 4, and 5 89

4

Use polynomial long division to rewrite f (x) in “mixed number” form.

23 2 3( )

3 4

x xf x

x

2

2

2

3 4 3 2 3

(3 4 )

6 3

(6 8)

5

x

x x x

x x

x

x

5( ) 2

3 4f x x

x

Given that 3 is a zero of the cubic,find the other zeros. f (x) = 5x3 – 13x2 – 3x – 9

Then factoring the leftover quadratic: 5x2 + 2x + 3 …using the quadratic formula

Using synthetic division, 3 5 -13 -3 -9 15 6 9 5 2 3 0

22 2 4(5)(3) 2 4 60 2 56

2(5) 10 10

2 1 4 14 2 2 14 1 14

10 10 5

x

i i

Decide the degree, and number of realand non-real zeros of the polynomial.

There are 6 branches(increase/decrease parts)so it is sixth degree, and there are two real zeros,and 4 non-real zeros.

2

-2

-4

5

For the rational function, find the coordinates of the hole.

The factor that will cancel from top and bottom is the (x + 1).So you know the hole is at the x-coordinate of –1. Now to find the y-coordinate… Plug x = -1 into what is left when you cancel that factor.

( 2)( 1)( 4)( )

( 1)( 3)( 4)

x x xf x

x x x

( 1 2)( 1 4) ( 3)(3) 9( 1)

( 1 3)( 1 4) ( 4)( 5) 20f

Hole: (-1, - 9/20)

Find the vertical asymptotes of the rational function:

Vertical asymptotes happen when the denominator only is zero. The 2x + 1 factor cancels, so it is a hole and NOTa vertical asymptote. The other two factors in the denominator show where the vertical asymptotes are.

(2 1)(3 2)( 5)( )

( 5)(2 1)(2 3)

x x xf x

x x x

x – 5 = 0 and 2x + 3 = 0So solving these we getx = 5, and x = - 3/2

2

2

2

2

9 2 3lim

3 4 1

9 3lim lim 3

3 1

x

x x

x x

x x

x

x

Find the horizontal asymptote of the rational function:

2

2

9 2 3( )

3 4 1

x xf x

x x

The horizontal asymptote happens as x goes to infinity. As x gets big, the first terms on top and bottom dominate over the others.

So only look at those that are dominating, and we see that they simplify…

So the horizontal asymptote is y = 3.

Break the rational function into partial fractions.

Beginning we break it into …

5 2( )

( 2)( 4)

xf x

x x

5 2

( 2)( 4) 2 4

x A B

x x x x

Then making the common denominator …

5 2 ( 4) ( 2)

( 2)( 4) ( 2)( 4) ( 4)( 2)

x A x B x

x x x x x x

The numerators give the equation … 5 2 ( 4) ( 2)x A x B x

If x = 4, then the equation is … 18 (0) (6) 3A B B

If x = - 2, then the equation is … 12 ( 6) (0) 2A B A

So, plugging in these values, we get…

2 3( )

( 2) ( 4)f x

x x

Find the coordinates of any hole(s) and vertical asymptote(s).

The factor of x + 3 cancels, so there is a hole at –3.

( 3)(2 1)(4 3)( )

(4 3)( 1)( 3)

x x xf x

x x x

( 7)( 9) 63 21( 3)

( 15)( 4) 60 20f

Hole at (–3, 21/20).

The factors left in the denominator help find the vertical asymptotes at x = ¾ and x = 1.

Identify the type of conic section.3x2 + 9y2 – 36x + 36y – 192 = 0

Since the x2 and y2 terms are both positive, but

have different coefficients, it is an ellipse.

Find the Cartesian Equation (showing transformations) :

The center is at (3,-2),the x-dilation is 2, and the y-dilation is 5. So the equation is …

5

-5

-10

10

2 23 2

12 5

x y

Transform the parametric equations into a Cartesian equation for the conic.

For this, we solve each equation for cosine or sine, and plug them into a Trig. Pythagorean Property.

4 5cos

2 3sin

x t

y t

2 2

2 2

44 5cos 4 5cos cos

52

2 3sin 2 3sin sin3

cos sin 1

4 21

5 3

xx t x t t

yy t y t t

t t

x y

Transform the given Cartesian equationinto the general Cartesian equation.

This is where we need expand …

2 25 2

13 4

x y

2 2 2 2

2 22 2

2 2

2 2

2 2

5 2 ( 5) ( 2)1 1

3 4 9 16

multiply both sides by 144...

( 5) ( 2)144 144 16( 5) 9( 2) 144

9 16

16( 10 25) 9( 4 4) 144

16 160 400 9 36 36 144

16 9 160 36

x y x y

x yx y

x x y y

x x y y

x y x

292 0y

Transform the general Cartesian equationinto the Cartesian equation that shows transformations.4x2 + 16y2 + 16x – 160y + 352 = 0This is where we need to complete the square …

2 2

2 2

2 2

2 2

2 2 2 2

2

4 16 16 160 352

4( 4 ___) 16( 10 ___) 352

4( 4 4) 16( 10 25) 352 4 4 16 25

4( 2) 16( 5) 64 Divide both sides by 64....

4( 2) 16( 5) 64 ( 2) ( 5)1

64 64 64 16 4

2 5

4 2

x x y y

x x y y

x x y y

x y

x y x y

x y

2

1

Change the polar point (r, θ) = (24, 210°) to a Cartesian point (x, y). Exact answer please.

Use the conversion formulas: x = r cos θ and y = r sin θ

x = 24 cos 210° = 24(-√3/2) = -12√3y = 24 sin 210° = 24(-1/2) = -12

(x, y) = (-12√3, -12)

Find the polar equation for the given graph.

4

2

2

4

5 5

It is a rose with 5 leaves. All roses are of the form… r = a cos (nθ) or r = a sin (nθ). Since this graph is symmetric about the y-axis, it is a sine equation. Since it has an odd number of leaves, then n = number of leaves. The farthest out the graph goes is 5, so a = 5.

SO, the equation is r = 5 sin (5θ).

Divide these complex numbers in polar form.40cis200

5cis10

40cis200 40cis (200 10 ) 8cis190

5cis10 5

Use De Moivre’s Theorem to evaluate.

1

532cis 200

11

1555

(with 5 evenly spaced solutions) 360 / 5 72

32cis200 (32) cis ( 200 )

2cis40

2cis112

2cis184

2cis256

2cis328

+ 72°

+ 72°

+ 72°

+ 72°

Transform the polar equation into a Cartesian equation that shows the transformations.

16sinr

2 22

2 22 2

2 22 2

2 2

16sin( 8) 64

16 sin( 8)

16 164 64

16 08

116 64 0 64 8 8

rx y

r rx y

x y y

x y yx y

x y y