Adventures in Precalculus

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A practical tutorial on precalculus for high school students who have completed Algebra 1.;useful for self-learning and also as review for starting calculus course...many advanc d concepts are a simple manner; The tutorial has several simple examples and exercise problems to practise...Several applications are given in each topic, including exercise using graphing uitlity for polar equations.

Text of Adventures in Precalculus

Adventures in Precalculus - 1Dr N K Srinivasan

What is Precalculus?If you scan through a 'Precalculus' book, you will note that it is a course with a jumble of topics--more like a bridge course from Algebra 2 to regular Calculus. But a careful study will show you something: It teaches you several mathematical functions and how to manipulate them and use them for 'modeling' real-world problems.

Well, you have to learn many mathematical functions because these functions are the 'ingredients' for using Calculus.While learning these functions, you will use a bag of clever tricks,call them 'tools of the trade" if you like. This article teaches you some of the clever tricks you will use--so that 'precalculus' becomes easy and interesting to study and develops your 'calculus muscles' to handle tougher math problems later.! Note: Get a graphing utility and use as required.

1 Completing the square:

Take a simple example: Solve

x +6x -16 =02

You can, of course, use the quadratic formula ,if you remember it ! Is there a simple way to solve? Recall : 2ax + a2



= (x+a)(x+a) = x



Keep this equation in mind. We will proceed from right to left while "completing the square". Take the example and rewrite this equation as follows:


+ 6x

= 16

Take the coefficient of x, which is 6 in this example; divide this by 2 and

square the number: that is 6/2=3 and 3 x3 =9 Add this number both sides:+9 You can factor the left side using the relation given earlier:



+ 6x +9 = 16

(x + 3)2

= 25

Take the square root on both sides: x+3 = +/- 5 So the roots are: x+3 =5 or x=2 or

and x+3 = -5 x=-8

This method is far easier than using quadratic formula.

Try the following problems before you proceed: 1 solve x2

+ 4x -32 =0

[Ans ; x=

4, x=-8 2 Solve x - 8x -9=02


x=9;x=-1] 3 Solve 3x +6x -24 =02




Example 2 This method--"completing the square has other uses too. We will see one application: Solve x2

+ y +2x +6y -6 =02

This problem appears 'tricky'...well -this is a 'standard' equation for a circle. Recall that the equation for a circle with centre C(0,0) at the origin is : x + y = r2 2 2

where r is the radius.

The equation changes when the center is shifted to C'(h,k) position.

The equation becomes:


2 + (y-k)2 = r2

Now the given equation can be factored by "completing

the square" for both x and y . Rewrite the expression:


+ 2x + ) + ( y2+6y + ) = -6

Add 1 to the first bracket and 9 to the second bracket and the same numbers on the right side



2 + 2x +1) + (y2 + 6y + 9) = -6 +1+9

Factoring for x and y we get:

(x+1)2 + (y+3)2 = 4

The problem is solved : this is a circle with centre at C(-

1,-3) and radius 2.

Problem to solve:

1 find the center and radius of the circle: x2 + y2 -4x


[Ans center C(2.-5) radius 6]

2 Find the center and radius of this circle: x2

+x + y2 -6y +1/4 =0

[Ans center (-1/2,3), radius 3]

Example 3 Find the vertex of the parabola whose equation is:


-6x-y +10 =0

Let us rewrite this first: )+10 = y

(x -6x2

The coefficient of x is 6; so let us put 9 inside the bracket and subtract 9 outside: ( x +10 =y (x -3) 1 This is a parabola, with vertex at [3,1] Try this:2 2

-6x +9) -9

= y -

1 Find the vertex of the parabola : 2x2

-8x -y + 7 =0

We can extend this method to find the equation of an ellipse from

the equation in

standard form:Read this from your text book .

2 Function of a functionSuppose you have a function: y= f(x) = 2x +3

You have another function : z = g(x) = x2


you can generate another function f [ g(x)] Write this as: f[ g(x)] = f [x +1]2

Then write the function f(x) replacing x by x2

+ 1.2

We get f[g(x)] = 2(x +1) + 3

You also get f[f(x)] by writing ;

f [ 2x+3] = 2 (2x+3) + 3 = 4x + 9Try writing g[g (x)] Example: g[f(x)] Are they same? no. Given f(x) = 1/x + 3. g(x)= x -1. write f[g(x)] and

Complex formulae can be created by using this method:Example 1: The electrical resistance of a wire increases with temperature as follows: R = a + bT + c T2Now the temperature of an electrical heater varies with time t as follows:

T = d + 4 t Find the rate of heating by using the relation : Heat produced H = I2

R as a function of time:

R =

f(T) = a + bT + c T.T

T (t) = d + 4t So f [ T (t)] = f [ d+4t] = a + b

[d+4t] + c [ d+4t][d+4t] Now you see the usefulness of function of function. Here is another example: example 2: The atmospheric air pressure decreases almost linearly [like a straight line] as we ascend to about 10000 feet.The pressure P = P0 2 (h) where h is height in feet.

A baloon ascends at the rate of 50 feet per minute. where t is time in minutes from the ground level.

h = 50 t

Write the equation for change in pressure with respect to time for a traveller going up in the balloon.

3 Inverse functionsTake the function y = f(x) = 2x + 3 You can find the inverse function by switching x and y: Replace x by y abd y by x: Now solve for Y : x= 2y + 3 2y = x - 3 y= (x-3)/2 Note that this function is an inverse function of f(x) . Let us write this as f-1 (x)

Note that the inverse function is not the reciprocal of f(x); it is not 1/f(x). The inverse function is the reflection to the original function across the line y=x. Y=x is the diagonal line that passes through the origin.

Draw the lines y= 2x +3 and the line y=(x-3)/2 in a graph paper. You will see the two lines as mirror images of each other. Example 2 Find the inverse function for the parabola: y = 4 x.x Switching x and y ; x= 4 y.y

or y= sqrt(x)/2 Plot again ,using your graphing utility, the two curves: y= 4xx and y= sqrt(x)/2 .Do they form reflection across the diagonal line y=x? Check this.

What if we find f[ f-1

(x)] ?-1

Take the example of f(x)= 2x+3 , and f (x) = (x-3)/2 f[ (x-3)/2] = 2[(x-3)/2] + 3 = x-3 +3 =x

Are you surprised with the result? We took x, transformed it into f(x) ,then took its inverse and again revert it back by f(x) ,getting x again. So, you can always check your result of inverse function by forming f [ f-1

(x)]----if you get x, you are right.Exponential function and logarithmic function are inverses of each other. y =f(x)= e Switch y and x:x

x = ey

Solve for y:

ln x = y= f-1


So the two functions are inverses. Plot the two functions using your graphing utility; you will find that they are curves reflected over y=x. Try these problems: 1 If f(x) = sqrt(2x+3), find its inverse function. 2 If f(x) = 2ex find its inverse function.

ApplicationWe can use inverse functions to find the conversion formula for temperature in Centigrade scale to Fahrenheit and the reverse one: If x is the temp in centigrade and y the temp in Fahrenheit, we have the relation as follows: y = 32 + (9/5) x Let us find the inverse function; switch x and y in the above equation: x = 32 + (9/5) y Solve for y: y = (x-32) (5/9)

This formula gives the conversion fromula when x is in Fraherheit and y the centigrade.! Example: If 212 F is the boiling point of water, find the temperature in centigrade scale: deg C = (212-32) (5/9) = (180 x 5 )/9 = 100 deg C So, the boiling point of water is 212 deg F or 100 deg C.

3 Shifting and scalingMany functions can be shifted around and scaled up or down [multiplied or divided] by suitable manipulation. Consider a parabola with vertex at the origin V(0,0): y = x2Suppose you want to shift the vertex to V(2,0) , change x to x-2. y = (x-2)2 You want to shift the vertex to V(2,1), change y to y-1. y-1 = (x-2)2Consider a circle with center at the origin C(0,0) and radius 2 units>

The equation for the cirlce is :

x + y = 42 2

Suppose you want to shift the center of the circle to P(-2,1) with the same radius,


+ (y-1)




Another transformation you do is to enlarge or shrink by multiplying or dividng or y.

y = k x.x is a parabola that rises very fast along the y axis--if k is greater than 1.

Take y = 2 x2Compare this parabola with y = x2 We call this process "scaling" Again consider the graph of y = You can choose k=1, k=2 k=1/2 shapes you get. 1 Write the equation for a ellipse with centre at (-2,2) k . Draw the graphs and study the

2 Write the equation for a cirlce with center at (2,-2) and radius 3 units.

4 parametric equationsWhen we relate x to y through the function , y = f(x) , we have a single graph of y versus x. Often it is convenient to use another variable which is intermediate and common to both: Let us call this intermediate variable t; then x = f(t) and y= g(t) We use two functions now, both functions of t. Consider a sphere whose surface area is related to radius: A = f (r) = 4 .pi.r.r the volume is related to radius: V = g(r)= 4.pi.r.r.r/3

Here the radius is the common parameter. We can take the ratio of surface area to volume: A/V = f(r)/g(r)= 3/r So writing parametric equations help us to see new relations.Thus if you want to study cooling of a sphere [like our Earth] we want to study the ratio of surface area to

volume.As r increases, the ratio of surface area to volume decreases, so the cooling rate could get lower. If your housetop is like a hemispherical dome, it would cool slo