Precalculus requires a thorough understanding of functions and the language used to work with functions.
The web exercises below will give you the required review.
Be sure to clickclickclick through some of the exercises
in each of these sections! The sections will open in a new tab/window.
Then, do the web exercise on
this page to practice with difference quotients, which are particularly important in calculus.
EXAMPLE:
difference quotients
Consider a point $\,\bigl(x,f(x)\bigr)\,$ on the graph of a function $\,f\,$.
If $\,h\,$ is a small positive number, then $\,x+h\,$ lies a little to the right of $\,x\,$.
If $\,h\,$ is a small negative number, then $\,x+h\,$ lies a little to the left of $\,x\,$.
In both cases, $\,\bigl(x+h,f(x+h)\bigr)\,$ is a point on the graph of $\,f\,$, likely close to the original point $\,(x,f(x))\,$.
The slope of the line through $\,\bigl(x,f(x)\bigr)\,$ and $\,\bigl(x+h,f(x+h)\bigr)\,$ is:
$$\text{slope} = \frac{\text{change in $y$}}{\text{change in $x$}}
= \frac{f(x+h)  f(x)}{(x+h)x} = \frac{f(x+h)f(x)}{h}$$

To slide the point $\,\bigl(x+h,f(x+h)\bigr)\,$
closer to
the original point $\,\bigl(x,f(x)\bigr)\,$,
To keep the graph uncluttered,
the labels for the new (closer) points
$\bigl(x+h,f(x+h)\bigr)$
are not shown.


Recall that the result of a division problem, $\,\frac{a}{b}\,$, is called a quotient,
and the result of a subtraction problem, $\,a  b\,$, is called a difference.
The expression $\displaystyle\,\frac{f(x+h)f(x)}{h}\,$ is called a difference quotient,
because it is a quotient of differences:
 the numerator, $\,f(x+h)f(x)\,$, is a difference
 the denominator, $\,h = (x + h)  x\,$, is a difference (in disguise)
 the entire expression is the quotient of these two differences
In calculus, you often have to simplify difference quotients, trying to rename in a way that
eliminates the $\,h\,$ in the denominator.
For example, if $\,f(x) = x^2 + 2x + 1\,$, then
$$
\eqalign{
\frac{f(x+h)  f(x)}h &= \frac{\overbrace{(x+h)^2 + 2(x+h) + 1}^{f(x+h)}  (\overbrace{\vphantom{(x+h)^2} x^2 + 2x + 1}^{f(x)})}h \qquad& \text{function evaluation} \cr
&= \frac{x^2 + 2xh + h^2 + 2x + 2h + 1  x^2  2x  1}{h} \qquad& \text{FOIL; distributive law}\cr
&= \frac{2xh + h^2 + 2h}{h} &x^2  x^2 = 0;\ \ 2x  2x = 0;\ \ 1  1 = 0\cr
&= \frac{h(2x + h + 2)}{h} = \frac hh\cdot (2x + h + 2) &\text{factor out the $h$}\cr
&= 2x + h + 2 &\text{cancel: } \frac{h}{h} = 1\cr
}
$$
Note that the final form, $\,2x + h + 2\,$, has no $\,h\,$ in the denominator.
Note also that as $\,h\,$ gets closer and closer to $\,0\,$, the expression $\,2x + h + 2\,$
gets closer and closer to $\,2x + 2\,$.
You'll learn in calculus that $\,2x+2\,$ is called the ‘
derivative of $f\,$’ and is denoted
by $\,f'(x)\,$ (read as ‘$f$ prime of $\,x\,$’).
This new function $\ f'(x) = 2x + 2\ $ gives the slope of the tangent line to the graph of $\,f\,$ at the point $\,\bigl(x,f(x)\bigr)\,$!
Since $\,x\,$ can represent
any real number, you might prefer to say:
This new function $\ f'(x) = 2x + 2\ $ gives the slope
s of the tangent line
s to the graph of $\,f\,$ at the point
s $\,\bigl(x,f(x)\bigr)\,$!
NOTE: The notation ‘$\,\Delta x\,$’ (read aloud as ‘delta $x$’)
is often used instead of $\,h\,$ to denote a small change in $\,x\,$.
With this notation, the slope of the line through the point $\,\bigl(x,f(x)\bigr)\,$ and nearby point $\,\bigl(x+\Delta x,f(x+\Delta x)\bigr)\,$
is given by the difference quotient:
$$\frac{f(x+\Delta x)  f(x)}{\Delta x}$$
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.