graphing and functions one of my problem says that the average slope formula is: $\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$
It tells us the average slope over the interval from $x_1$ to $x_2$
Then it says to take the average slope of the function $x^2+3x+1$ over one unit intervals from $0$ to $10$. 
can i use a particular function function to solve this problem?
 A: As André said, what they want is the 10 numbers
$$\frac{f(1)-f(0)}{1-0}, \frac{f(2)-f(1)}{2-1}, \ldots, \frac{f(10)-f(9)}{10-9}$$
-- that is, with $x_1$ increasing from $0$ to $9$ while $x_2$ increases from $1$ to $10$. Then plot them against the $x_1$ values for each of them, and look for a pattern.
A: Since each interval is of length $1$ you could look at $$\dfrac{f(x+1)-f(x)}{(x+1)-(x)} = \dfrac{((x+1)^2+3(x+1)+1)-(x^2+3x+1) }{(x+1)-(x)}$$ for $x=0,1,2,\ldots,9$.  You might find it easier if you simplified the expression. 
A: The naive approach is to make yourself a table, with $10$ rows (one for each $x_i$ with: $x_1 = 1, x_2 = 2, \dots, x_{10} = 10$). The first column would be $f(x_i)$, the second column would be $x_i - x_{i-1}$ (you should notice an obvious pattern in this column), the third column would be $f(x_i) - f(x_{i-1})$, and the fourth column would be:
$\dfrac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}$
(and if you noticed the pattern in the second column, you might not need this last one....why?)
To get this started, set $x_0 = 0$, with $f(x_0) = f(0) = 1$.
Note that Henry's answer above gives a short-cut, this is "the long way".
The first row (for $x_1 = 1$) would look like this:
$f(1) = 5;\ \ x_1 - x_0 = 1 -0 = 1;\ \ f(1) - f(0) = 5-1 = 4;$ slope $= 4/1 = 4$.
Nine more to go.... 
A: $f(x)=x^2+3x+1, x_1=0, x_2=10$
so the average slope on the interval [0,10] is $\frac{f(10)-f(0)}{10-0}=\frac{131-1}{10}=13$. 
If you want the average slope over all 1 unit intervals $[a, a+1]$ for $0\le a\le 9$ then this is the symbolic expression $\frac{f(a+1)-f(a)}{a+1-a}$ which simplifies to $2a+4$.
For example with $a=4.5$ we also get average slope of 13, but for $a=9$ we get average slope of 22.
