finite mathematics & applied calculus topic summary:
Nonlinear functions and models: Part 1 of 2
Tools:

A quadratic function is a function of the form
$f(x)=ax^2+bx+c \qquad (a \neq 0)$.
Its graph is called a parabola. The vertex of this parabola occurs at the point on the graph with $x$- coordinate
$x = -\frac{b}{2a} \qquad$ Vertex
It crosses the $y$-axis ($y$-intercept) at
$y = c \qquad$ y-Intercept
It crosses the $x$-axis ($x$-intercept(s)) at the solution(s) of the quadratic equation $ax^2+bx+c = 0$ (if there are any) given by
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. \qquad$ x-Intercept(s)
It is symmetric around the vertical line through the vertex.

If the coefficient $a$ of $x^2$ is positive, it is concave up (as in the figure below when you press "$a > 0$"). If $a$ is negative, it is concave down (as in the figure below when you press "$a < 0$").

Examples: Quadratic functions and their graphs

Parabola The curve
$y=x^2-2x-8$
is a parabola with vertex at
$x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1.$
The $y$-coordinate of the vertex is
$y = (1)^2 - 2(1) - 8 = -9.$
The $y$-intercept is $c = -8$ and the $x$-intercepts are the solutions of
$x^2 - 2x - 8 = 0$
$(x+2)(x-4) = 0$,
so that $x = -2$ and $4$. Here is its graph:

Practice:

Laws of exponents

If $b$ and $c$ are positive, and $x$ and $y$ are any real numbers, then the following laws hold:
 Law Example $b^xb^y=b^{x+y}$ $2^3 2^2=2^5 = 32$ $\frac{b^x}{b^y}=b^{x-y}$ $\frac{4^3}{4^2}=4^{1} = 4$ $\frac{1}{b^x}=b^{-x}$ $9^{-0.5}=\frac{1}{9^{0.5}} = \frac{1}{3}$ $b^0 = 1$ $3.1415^0 = 1$ $(b^x)^y = b^{xy}$ $(2^3)^2 = 2^6 = 64$ $(bc)^x = b^xc^x$ $(4 \cdot 2)^2 = 4^22^2 = 64$ $\left(\frac{b}{c}\right)^x = \frac{b^x}{c^x}$ $\left(\frac{4}{3}\right)^2 = \frac{4^2}{3^2} = \frac{16}{9}$
Example: Laws of exponents

Practice:

Exponential functions

An exponential function is a function of the form
$f(x) = Ab^x$,
where $A$ and $b$ are constants and $b \gt 0$.

The number $b$ is called the base of the exponential function.

Role of b
• An increase of 1 unit in $x$ has the effect of multiplying $f(x)$ by $b$.
• An increase of 2 units in $x$ has the effect of multiplying $f(x)$ by $b^2$.
• ...
• In general, an increase of $\Delta x$ units in $x$ has the effect of multiplying $f(x)$ by $b^{\Delta x}$.

Role of A
• $f(0) = A$, so $A$ is the $y$-intercept of the graph of $f$.
Examples: Exponential functions

The function $f(x) = 3(2^x)$ is an exponential function with $A = 3$ and $b = 2$. Some values of $f$ are shown in the following table:
Its graph is as follows:

Practice:

Compound interest

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
$F(t) = P\left$1+\frac{r}{m}\right$^{mt}$.
This is an exponential function of $t$ of the form $F(t) = Ab^t$ as it can be written as
$F(t) = P\left(\left$1+\frac{r}{m}\right$^{m}\right)^t \qquad$ $A = P, \quad b = \left$1+\frac{r}{m}\right$^{m}$
Examples: Compound interest

You invest \$1000 at an annual rate of 4.8% interest, compounded monthly. This means that$P = 1\,000, \quad r = 0.048, \quad m = 12.$Substituting gives $F(t)= 1\,000\left(1+\frac{0.048}{12}\right)^{12t}= 1\,000(1.004)^{12y}$This function gives the value of the investment after$t$years. For instance, after 5 years, the investment is worth$F(5) = 10\,000(1.004)^{12 \times 5} = 10\,000(1.004)^{60} = $\$12,706.41.

The number e

The numbers
$\left(1+\frac{1}{m}\right)^m$.
converge to the number e = 2.71828182845904523536... as $m$ gets larger and larger. The following table shows the value of $\left(1+\frac{1}{m}\right)^m$ for several values of $m$.
Example: The number e

Enter your own value of $m$ and press "Compute". (You will notice that values of $m$ larger than around 100,000,000 will result in computational errors --- Experiment!).
 $m =$ $\left(1+\frac{1}{m}\right)^m =$

Continuous compounding

The number $e$ appears in the formula for continuous compounding: If an amount $P$ (the present value) earns interest at an annual interest rate $r$, compounded continuously, then the accumulated amount (or future value) after $t$ years is
$F(t) = Pe^{rt}$.
The effective (annual) yield from continuous compounding is given by
$r_e = e^r - 1$.
Examples: Continuous compounding

If \$10,000 is invested at an annual interest rate of 4.8% compounded continuously, then the accumulated amount after$t$years is$A = 10\,000e^{0.048t}$The effective annual yield is$r_e = e^{0.048} - 1 \approx 0.04917$, that is, 4.917% per year. Logarithms The statement:$\log_bx = y$The base b logarithm of x equals y means$b^y = x.$Note: •$\log_{10}x$is often written as$\log x$, and called the common logarithm of$x.$• The expression$log_ex$is often written as$\ln x$and called the natural logarithm of$x.$Examples: Logarithms The following table lists some exponential equations and their equivalent logarithmic form: Practice: Logarithm identities The following identities hold for any positive base$b \neq 1$and any positive numbers$x$and$y.$ Identity Example$\log_b(xy) = \log_bx + \log_by\log_2 16 = \log_2 8 + \log_2 2\log_b\left(\frac{x}{y}\right) = \log_bx - \log_by\log_3\left(\frac{5}{4}\right) = \log_3 5 - \log_3 4\log_b(x^r) = r\log_bx\log_4(6^5) = 5\log_4 6\log_bb = 1$$\log_b 1 = 0 \log_2 2 = 1$$\log_3 1 = 0\log_b\left(\frac{1}{x}\right) = -\log_b x\log_5\left(\frac{1}{4}\right) = -\log_5 4\log_bx = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}\log_25 = \frac{\log 5}{\log 2} \approx 2.3219\$
Example: Logarithm identities

Practice: