# How do you find the slope of the secant lines of  f(x) = x^2 + 5x at (6 , f(6)) and (6 + h , f(6 + h))?

Jan 23, 2017

Slope $= h + 17$

#### Explanation:

We have $f \left(x\right) = {x}^{2} + 5 x$

When $x = 6$:

$f \left(x\right) = {6}^{2} + 5 \cdot 6$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 36 + 30$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 66$

When $x = 6 + h$:

$f \left(x\right) = {\left(6 + h\right)}^{2} + 5 \left(6 + h\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 36 + 12 h + {h}^{2} + 30 + 5 h$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {h}^{2} + 17 h + 66$

So the slope of the secant line at $\left(6 , f \left(6\right)\right)$ and $\left(6 + h , f \left(6 + h\right)\right)$ is;

$\frac{\Delta y}{\Delta x} = \frac{f \left(6 + h\right) - f \left(6\right)}{\left(6 + h\right) - 6}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{{h}^{2} + 17 h + 66 - 66}{h}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{{h}^{2} + 17 h}{h}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = h + 17$

Side Note - What is the significance of this:
If we let $h \rightarrow 0$ then the secant line becomes the tangent line, and that tangent would therefore have slope $17$ when $x = 6$.

With our knowledge of Calculus we can confirm this as:

$f ' \left(x\right) = 2 x + 5 \implies f ' \left(6\right) = 12 + 5 = 17$