A quadratic function is a function of the form

f(x) = ax² + bx + c     (with a ≠ 0).

Its graph is called a parabola.

The vertex of this parabola occurs at the point on the graph with x coordinate -b/(2a).

It crosses the y-axis (y-intercept) at y = c.

It crosses the x-axis (x-intercept(s)) at the solutions of the quadratic equation ax² + bx + c = 0 (if there are any).

It is symmetric around the vertical line through the vertex.

If the coefficient (a) of x² is positive, it is concave up (as in the example to the right). If a is negative, it is concave down (as in the figure below).

The parabola

y = x² - 2x - 8

-

b 2a

=

2 2

=

1.

y = (1)² -2(1) - 8 = -9.

x² - 2x - 8 = 0
(x + 2)(x - 4) = 0,

An exponential function is a function of the form

f(x) = Ab^x,

The function f(x) = 3(2^x) is an exponential function with A = 3 and b = 2. It has the following graph.

The following table shows the y-coordinates of points on this graph. All you do is supply the x-coordinates and press "Compute y"

If b and c are positive, and x and y are any real numbers, then the following laws hold.

Law	Example
bx by = bx+y	23 22 = 25 = 32
bx by = bx-y	43 42 = 41 = 4
1 bx = b-x	1 90.5 = 9-0.5 = 1 3
b0 = 1	(3.3)0 = 1
(bx)y = bxy	(32)2 = 34 = 81
(bc)x = bx cx	(4 2)2 = 4222 = 64
b c x = bx cx	4 3 2 = 42 32 = 16 9

Future Value
If an amount P (the present value) earns interest at an annual interest rate r, compounded m times per year, then the accumulated amount (or future value) after t years is

A

=

P

1

+

r m

mt

A(t)

=

P

1

+

r m

mt

You invest $1000 at an annual rate of 4.8% interest, compounded monthly. This means that

P = 1000,     r = 0.048,     m = 12.

A(t)	=	1,000 1 + 0.048 12 12t
	=	1,000(1.004)12t	.

A(5) = 1000(1.004)¹²⁵ = 1000(1.004) ⁶⁰ = $1270.64.

The numbers

1

+

1 m

m

m	1	10	100	1000	10000
1 + 1 m m	2	2.59374246	2.70481383	2.71692393	2.71814593

The number e appears in the formula for continuous compounding: If $P is invested at an annual interest rate r compounded continuously, then the accumulated amount after t years is

A = Pe^rt.

r_e = e^r - 1.

If $1000 is invested at an annual interest rate of 4.8% compounded continuously, then the accumulated amount after t years is

A = 1000e^0.048t.

r_e = e^0.048 - 1 ≈ 0.04917,

The statement

log_ax = y

a^y = x.

The following table lists some exponential equations and their equivalent logarithmic form.

Exponential Form	103 = 1000	42 = 16	51 = 5	70 = 1	4-2 = 1/16
Logarithm Form	log1000 = 3	log416 = 2	log55 = 1	log71 = 0	log4(1/16) =-2

The following identities hold for any positive a 1 and any positive numbers x and y.

Identity	Example
loga(xy) = logax + logay	log216 = log28 + log22
loga(x/y) = logax - logay	log2 (5/3) = log25 - log23
loga(xr) = r logax	log2(65) = 5 log26
logaa = 1 loga1 = 0	log22 = 1 log31 = 0
loga(1/x) = -logax	log2(1/3) = -log23
logax = logx loga = ln x lna	log25 = log 5 log 2 ≈ 2.3219

If a is any positive number, then the functions f(x) = log_ax and g(x) = a^x are inverse functions. This means that

a^logax= x

log_a(a^x) = x

Want to learn more about inverse functions? Go to the on-line text on inverse functions.

2^log2x = x

e^{ln x} = x

log₂(2^x) = x

ln (e^x) = x

An exponential decay function has the form

Q(t) = Q₀e^-kt           Q₀, k both positive

The half-life t_h of a substance undergoing exponential decay is the amount of time it takes for half the original quantity to decay. The half-life does not depend on the original quantity of substance present.

The decay constant k and half-life t_h for Q are related by

t_h k = ln 2.

Exponential Growth and Doubling Time

An exponential growth function has the form

Q(t) = Q₀e^kt           Q₀, k both positive

The doubling time t_d of a substance undergoing exponential growth is the amount of time it takes for half the original quantity to double. The doubling time does not depend on the original quantity of substance present.

The growth constant k and doubling time t_d for Q are related by

t_d k = ln 2.

Decay and Half-Life:

1. Q(t) = Q₀e^{-0.000 120 968t} is the decay function for carbon-14.

2. If k = 0.0123, then t_h(0.0123) = ln 2, so the half-life is t_h = (ln 2)/k = (ln 2)/0.0123 ≈ 56.35 years.

Growth and Doubling Time:

1. P(t) = 1000e^0.5t is the amount of money in an account after t years if $1000 invested at 5% annually with interest compounded continuously.

2. If k = 0.0123, then t_d(0.0123) = ln 2, so the doubling time is t_d = (ln 2)/k = (ln 2)/0.0123 ≈ 56.35 years.

A logistic function has the form

f(x) =

N 1 + Ab-x

(A, N, b constant, b positive and ≠ 1)

Properties of the Logistic Curve

• The graph is an S-shaped curve sandwiched between the horizontal lines y = 0 and y = N. N is called the limiting value of the logistic curve.
• If b > 1 the graph rises; if b < 1, the graph falls.
• The y intercept is N/(1 + A)
• Role of b: For small x, the logistic function grows approximately exponentially with base b, and follows the curve [N/(1+A)] b^x.

b > 1

0 < b < 1

N = 50, A = 24, b = 3 gives

f(x) = 50 1 + 24(3-x)

Technology format: 50/(1+24*3^(-x))

The following figure shows the graph of f together with the exponential approximation

Logistic curve: 50/(1+24*3^(-x))
Exponential curve: 2*3^x