# How do you find the slope of the secant lines of f (x) = 2x^2 - 3x - 5 through the points: (2, f (2)) and (2 + h, f (2 + h)?

Nov 2, 2015

Find the slope of the line through the two points.

#### Explanation:

Slope of a line through points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{\Delta y}{\Delta x}$

We will need $f \left(x\right) = 2 {x}^{2} - 3 x - 5$
to find the $y$ values of the points: $\left(2 , f \left(2\right)\right)$ and $\left(2 + h , f \left(2 + h\right)\right)$

$m = \frac{f \left(2 + h\right) - f \left(2\right)}{\left(2 + h\right) - \left(2\right)}$

$= \frac{{\overbrace{\left[2 {\left(2 + h\right)}^{2} - 3 \left(2 + h\right) - 5\right]}}^{f} \left(2 + h\right) - {\overbrace{\left[2 {\left(2\right)}^{2} - 3 \left(2\right) - 5\right]}}^{f} \left(2\right)}{h}$

$= \frac{\left[2 \left(4 + 4 h + {h}^{2}\right) - 3 \left(2 + h\right) - 5\right] - \left[2 \left(4\right) - 3 \left(2\right) - 5\right]}{h}$

(All we did was the squares in each bracket. Next we will distribute as needed.)

$= \frac{8 + 8 h + 2 {h}^{2} - 6 - 3 h - 5 - 8 + 6 + 5}{h}$

(We did the multiplication in the $f \left(2\right)$ part and also distributed the minus sign through the $f \left(2\right)$ part. Now we can see that some of this stuff adds to $0$.

$= \frac{\textcolor{red}{8} + 8 h + 2 {h}^{2} \textcolor{g r e e n}{- 6} - 3 h \textcolor{b l u e}{- 5} \textcolor{red}{- 8} \textcolor{g r e e n}{+ 6} \textcolor{b l u e}{+ 5}}{h}$

$= \frac{8 h + 2 {h}^{2} - 3 h}{h}$

$= \frac{2 {h}^{2} + 5 h}{h} = \frac{h \left(2 h + 5\right)}{h}$

$= 2 h + 5$ $\text{ }$ for $h \ne 0$

We exclude $h = 0$ because if we try to use $0$ for $h$ the last line is $5$, but the previous lines are not numbers at all -- they are not defined.