SAT 8

144	PART II / MATH REVIEW

E X A M P L E :

Which of the following graphs represent(s) a constant function?

I.	y	II.	y	III.	y

x

x

x

(A)I only

(B)II only

(C)III only

(D)I and II

(E)II and III

IIIis not a function because it does not pass the vertical line test. I represents the constant function, y = c. II represents the zero function, y = 0, but the zero function is also a constant function.

D is the correct answer.

(Answer)

DETERMINING THE MAXIMUM OR MINIMUM

The graph of a function can be increasing, decreasing, or constant for intervals of its domain. A function is

•decreasing, if the y value decreases from left to right,

•increasing, if the y value increases from left to right, or

•constant, if the y value remains unchanged from left to right.*

Take parabola y = x2, for example. The graph decreases when y < 0 and increases when y > 0. The maximum or minimum value of a function is often the point where the function changes behavior from decreasing to increasing.

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145

The Maximum or Minimum of a Quadratic Function

Let f (x) = ax2 + bx + c where a ≠ 0.

If a < 0, the parabola is concave down and has a maximum value. If a > 0, the parabola is concave up and has a minimum value.

The maximum or minimum is f (x) when x = − 2ba . In other words, it’s the y value of the parabola’s vertex.

E X A M P L E :

Find the maximum value of the function f (x) = −4x2 + 3x + 1.

Notice a is less than zero, so the function does, in fact, have a maximum

	value. Solve for the x coordinate of the function’s vertex:
x = −				b	= −						3			=			3		.
															8
				2a					2(−4)
Now find the value of the function when x =																									3
																									8
		3						3			2					3
f				= −		4							+ 3								+ 1
		8						8								8
				= −			9	+				9	+ 1		=					(−9 + 18 + 16)		=	25	.
						16					8
																					16	16

The maximum value is	25	or 1.5625.	(Answer)
16
E X A M P L E :

A gardener has 60 feet of fencing to enclose a garden. Find the greatest possible area of the garden.

Let w = the width of the garden and l = the length of the garden.

2w + 2l = 60 w + l = 30

Write l in terms of w: l = 30 − w. The area of the garden is A = lw, so sub-

stitute 30 − w in for l.
A = (30 − w)w
Notice that this results in a quadratic equation.
A = 30w − w2 where a = −1 and b = 30.
The maximum value occurs when w = −	b	= −	30	= 15.
	2a		2(−1)

When w = 15, l = 30 − 15 = 15, so the area is 15(15) or 225 square feet.

225 square feet. (Answer)

146	PART II / MATH REVIEW

THE ROOTS OF A QUADRATIC FUNCTION

The roots, or solutions, of a quadratic equation are values of the variable that satisfy the equation. For example, factoring x2 + 2x − 15 = 0 results in (x + 5) (x − 3) = 0. x = −5 and x = 3 are the roots of the equation. When dealing with functions, the roots are the x values that result when f (x) = 0. You can also think of the roots as the x-intercepts of the graph of a function or the zeroes of the function.

If you’re given the roots of a quadratic equation, you can use them to determine the equation itself. A quadratic equation can be thought of as:

a[x2 (sum of the roots)x (product of the roots)] 0.

E X A M P L E :

The product of the roots of a quadratic equation is −14 and their sum is −3. Find a quadratic equation whose roots have the given product and sum.

Because the sum of the two roots is −3, and the product is −14, substitute these values into the equation.

a[x2 − (sum of the roots)x + (product of the roots)] = 0. a[x2 − (−3)x + (−14)] = 0.

Setting a equal to 1 results in one possible answer:

x2 + 3x − 14 = 0.

(Answer)

E X A M P L E :

Find a quadratic equation with integral coefficients having roots 6 and 2.

The sum of the roots is 8 and their product is 12.

a[x2 − (sum of the roots)x + (product of the roots)] = 0. a(x2 − 8x + 12) = 0.

Again, setting a equal to 1 results in one possible answer:

x2 − 8x + 12 = 0.

(Answer)

E X A M P L E :

Find a quadratic equation with integral coefficients having roots − 2 and 2.

The sum of the roots is 0, and their product is −2. a[x2 − 0x + (−2)] = 0.

Setting a equal to 1 results in one possible answer:

x2 − 2 = 0.

(Answer)

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147

E X A M P L E :

Find a quadratic equation with integral coefficients having roots 4 + i and 4 − i.

The sum of the roots is: 4 + i + 4 − i = 8.

The product of the roots is: (4 + i)(4 − i) = 16 − i2 = 17. a(x2 − 8x + 17) = 0.

Setting a equal to 1 results in one possible answer:

x2 − 8x + 17 = 0.

(Answer)

INVERSE FUNCTIONS

The inverse of a function is denoted by f –1 and satisfies the compositions f (f –1(x)) = x and f –1(f (x)) = x. All functions have an inverse, but the inverse is not necessarily a function (i.e., it doesn’t have to pass the Vertical Line Test.) A function has an inverse that is also a function, however, if the original function passes the Horizontal Line Test. This means that if every horizontal line intersects the graph of a function in at most one point, then the inverse of the original function is also a function.

If (a, b) is a point on the graph of f, then (b, a) is a point on the graph of f –1. In other words, the domain of f equals the range of f –1, and the range of f equals the domain of f –1. Because of this property, the graphs of f and f –1 are reflections of each other with respect to the line y = x. Some examples of functions and their inverse functions are as follows:

f (x) = x + 2. f −1(x) = x − 2.

g(x) = (x + 3) . g−1(x) = 2x − 3. 2

h(x) = x3 − 1. h−1(x) = 3 x + 1. f (x) = x2. no inverse function

g(x) = 1x . g−1(x) = 1x . h(x) = x. h−1(x) = x2.

On your graphing calculator, graph y1 = one of the given functions above, y2 = its inverse function, and y3 = x to see that the function and its inverse are reflections, or “mirror images,” of each other over the line y = x. Notice that the squaring function, f(x) = x2, does not have an inverse function because it doesn’t pass the Horizontal Line Test. The square root function, h(x) = x , does, however, have an inverse function because its domain is restricted to x ≥ 0.

To solve for the inverse of a function algebraically, interchange y and x, and solve for the new y value. This is the inverse function, if an inverse function

exists. For example, take f (x) =	x +	4	. Replacing f (x) by y results in: y =	x +	4	.
	2			2

Now interchange x and y and solve for y:

	148			PART II / MATH REVIEW
	x =		y + 4	.
		2
	2x = y + 4.
	2x − 4 = y.
	f −1	= 2x − 4.

Check your answer by graphing f and f −1 to see that they are reflections over the line y = x.

	y
	y = x
6
4	f (x)
2	f –1(x)
	x
–5	5
–2
–4
–6

E X A M P L E :

	If f (x) =					x + 6		, and f −1	is the inverse function of f, what is f −1(−1)?
		3
	f (x) =			x + 6			.
		3
	y =		x + 6		.
		3
	x =		y + 6		.
		3
	3x = y + 6.
	3x − 6 = y.
	f −1	= 3x − 6.
	f −1(−1) = 3(−1) − 6 = −9.									(Answer)
	An alternate solution is to set									x + 6	= −1. If (−1, n) is a point on the graph of
										3

the inverse function, then (n, −1) must be a point on the graph of the original function. By either method, f−1(−1) = −9.

CHAPTER 8 / FUNCTIONS	149
E X A M P L E :

If f (x) = 4x − 1, then what is f −1(x)?

Interchange the x and y values in the function, f (x) = 4x − 1 and solve for y.

y = 4x − 1.

x= 4 y − 1.

x+ 1 = 4y.

f (x)−1 =	x + 1.	(Answer)
	4

E X A M P L E :

If g(x) = x4, then what is g−1(x)?

y = x4 .

x= y4 .

±4 x = y. There is not a one-to-one correspondence between x and y values.

No inverse function.

(Answer)

RATIONAL FUNCTIONS

A rational function is defined by a rational (i.e., fractional) expression in one

variable and can be written as f (x) = p(x) . Unlike the linear and quadratic q(x)

functions that have been discussed thus far in this chapter, rational functions are not continuous.

They contain a break in the graph at the point where the denominator equals zero. (Limits of rational functions are discussed in the Numbers and Operations chapter.) The graph of a rational function f has vertical asymptotes at the zeroes of the denominator q(x).

E X A M P L E :

Find the domain of f (x) = x2 − 4 . x2 − 4x

The domain is the set of x values for which the function is defined, so

x2 − 4x cannot equal zero.
x2 − 4x = 0.
x(x − 4) = 0.
x = 0 or x = 4.
The domain is all real numbers except x = 0 and x = 4.	(Answer)

= x2 − 4

Note that the graph of f (x) x2 − 4x has vertical asymptotes at x = 0

and x = 4. Many graphing calculators do not handle asymptotes well, so you may get a graph that appears to be continuous with “zigzags” at x = 0 and x = 4. The graph should be discontinuous at x = 0 and x = 4 because the function is undefined at these domain values. On a TI

150	PART II / MATH REVIEW

graphing calculator, try changing the Mode to Dot, instead of Connected, to view the graph better.

E X A M P L E :

Find the zeroes of f (x)

= x3 − 3x

x2 − 9.

The zeroes of the function occur when f(x) = 0, or, in other words, at the

	x-intercepts of the graph.
0	=	x3	− 3x	.
		x2
			− 9
0 = x3 − 3x.					(x ≠ ±3).
0	= x(x2 − 3).
0	= x		or		x2 = 3
					x = ±	3.
The zeroes are: 0,−						3, 3 .	(Answer)
E X A M P L E :
What are the equations of the asymptotes of								f (x) =	x3	?
									x2 − 1

Vertical asymptotes occur at the zeroes of the denominator. x2 − 1 = 0

x = ± 1 = ±1

Note there are no horizontal asymptotes, since the degree of the numerator is greater than the degree of the denominator.

x = 1 and x = −1. (Answer)

HIGHER-DEGREE POLYNOMIAL FUNCTIONS

A polynomial function of x with degree n is given by: f (x) = an xn + an−1xn−1 + K + a2 x2 + a1x1 + a0 .

where n is a non-negative number, the coefficients of the x terms are real numbers, and an ≠ 0.

A first-degree polynomial function is a linear function: f (x) = ax + b (a ≠ 0). A second-degree polynomial function is a quadratic function: f (x) =

ax2 + bx + c (a ≠ 0).

The graphs of polynomial functions share the following properties.

1.They are continuous.

2.They have rounded curves.

3.If n (the highest exponent) is odd, and an > 0, the graph falls to the left and rises to the right.

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151

4.If n (the highest exponent) is odd, and an < 0, the graph rises to the left and falls to the right.

5.If n (the highest exponent) is even, and an > 0, the graph rises to the left and right.

6.If n (the highest exponent) is even, and an < 0, the graph falls to the left and right.

Properties 3 through 6 describe what is known as the Leading Coefficient Test, which describes the right and left behavior of the graphs of functions.

E X A M P L E :

Determine the right and left behavior of the graph of f (x) = −x3 + 7x.

The degree of the function is odd (n = 3) and its leading coefficient is −1. As x → −∞, y → ∞, and as x → ∞, y → −∞.

The graph rises to the left and falls to the right. (Answer)

The zeroes of a polynomial function are the x-values when f (x) = 0. A function of degree n has at most n real zeroes.

E X A M P L E :

Find the real zeroes of f (x) = x3 − 2x2 − 8x.

Because f (x) is a 3rd degree function, it can have, at most, 3 zeroes.

x3 − 2x2 − 8x = 0.
x(x2 − 2x − 8) = 0.
x(x − 4)(x + 2) = 0.
x = 0, x = 4, and x = −2.	(Answer)

E X A M P L E :

Find the real zeroes of f (x) = −x4 + 2x3 − x2.

Because f (x) is a 4th degree function, it can have, at most, 4 zeroes.

−x4 + 2x3 − x2 = 0.

−x2 (x2 − 2x + 1) = 0.

−x2 (x − 1)2 = 0.

x = 0 and x = 1.

(Answer)

Note that both x = 0 and x = 1 are repeated zeroes. Because n is even, the graph touches the x-axis at these points, but it does not cross the x-axis. (If n were odd, the graph would cross the x-axis at the repeated zeroes.)

Long division and synthetic division are also useful in factoring and finding the zeroes of polynomial functions.

152	PART II / MATH REVIEW

E X A M P L E :

Divide 2x3 − 9x2 + 7x + 6 by x − 2.

Let’s divide using synthetic division:

2	2	−9	7	6
		4	−10	−6
	2	−5	−3	0

The rightmost digit, 0, is the remainder. This means that x − 2 divides evenly into 2x3 − 9x2 + 7x.

The quotient is 2x2 − 5x − 3.

(Answer)

The Division Algorithm states that: f (x) = d(x) q(x) + r(x),

where d(x) is the divisor, q(x) is the quotient, and r(x) is the remainder. Applying this algorithm to the last example results in:

f (x) = (x − 2)(2x2 − 5x − 3).

You can further factor 2x2 − 5x − 3 to get: f (x) = (x − 2)(2x + 1)(x − 3).

Using synthetic division helps to factor polynomials that are not special products and easily factorable.

Some properties that are important in evaluating polynomial functions are:

1.The Remainder Theorem: If a polynomial f (x) is divided by x − r, then the remainder, r, equals f (r).

2.The Factor Theorem: A polynomial f (x) has a factor x − k if and only if f (k) = 0.

3.Descartes Rule of Signs: The number of positive real zeroes of a function is equal to the number of variations in sign of f (x) or less than that num-

ber by an even integer. The number of negative real zeroes of a function is equal to the number of variations in sign of f (−x) or less than that number by an even integer. (A variation in sign means that consecutive coefficients have opposite signs.)

4.Rational Root Test: If a polynomial function has integer coefficients, every

p

rational zero of the function has the form (simplified to lowest terms)

q

where p = a factor of the constant term a0, and q = a factor of the leading coefficient an.

5.Complex Zeroes Occur in Conjugate Pairs: If a polynomial function has real coefficients and a + bi (b ≠ 0) is a zero of the function, then a − bi is also a zero.

CHAPTER 8 / FUNCTIONS	153
E X A M P L E :

Using the rational root theorem, how many possible rational roots are there for x4 + 2x3 + 3x2 + 5x + 10 = 0?

The constant term is 10, so it can be factored as 1 × 10 or 2 × 5. The leading coefficient is 1, so it can only be factored as 1 × 1.

	p
Rational roots have the form			where p is a factor of 10, and q is a factor
	q
of 1, so possible factors are: ±1, ±2, ±5, ±10.
There are 8 possible rational roots.				(Answer)

E X A M P L E :

Find a polynomial function of lowest degree with real coefficients if two of its roots are 2 and 3 + i.

Because complex zeroes occur in conjugate pairs, 3 − i is also a zero of the function. Write the function as a product of its three factors, and then multiply to determine the function.

f (x) = (x − 2)[x − (3 + i)][x − (3 − i)]. f (x) = (x − 2)[(x − 3) + i][(3 − i) − i] f (x) = (x − 2)[(x − 3)2 − i2 ]

f (x) = (x − 2)(x2 − 6x + 9 − (−1))

f (x) = (x − 2)(x2 − 6x + 10)
f (x) = x3 − 8x2 + 22x − 20.	(Answer)

An alternative way to solve for the function is to recall that x2 − (sum of the roots)x + (product of the roots) = 0. Using 3 − i + 3 + i = 6 and (3 − i) (3 + i) = 10, you can quickly determine that the function is (x − 2)(x2 − 6x + 10) = x3 − 8x2 + 22x − 20.

E X A M P L E :

How many possible positive real roots does the function f (x) = 6x3 − 5x2 + 4x − 15 have?

Use Descartes Rule of Signs to determine the number of possible positive and negative real roots.

f (x) = 6x3 − 5x2 + 4x − 15 has 3 variations in sign, so there are 3 or 1 positive roots.

f (−x) = 6(−x)3 − 5(−x)2 + 4(−x) − 15 = = −6x3 − 5x2 − 4x − 15 has 0 variations in sign, so there are no negative roots.

1 or 3 positive real roots are possible.

(Answer)