# How do you find the slope of the secant lines of f (x) = x ^2 - 4x + 5  at (1, 2) and (5, 10)?

May 13, 2018

Slope of secant is $2$.

#### Explanation:

First let us confirm, whether $\left(1 , 2\right)$ and $\left(5 , 10\right)$ lies on $f \left(x\right) = {x}^{2} - 4 x + 5$.

As ${1}^{2} - 4 \cdot 1 + 5 = 1 - 4 + 5 = 2$, $\left(1 , 2\right)$ lies on the curve

and as ${5}^{2} - 4 \cdot 5 + 5 = 25 - 20 + 5 = 10$, $\left(5 , 10\right)$ lies on the curve.

Now slope of line joining $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is

$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

and hence slope of secant is $\frac{10 - 2}{5 - 1} = \frac{8}{4} = 2$

graph{(x^2-4x+5-y)((x-1)^2+(y-2)^2-0.01)((x-5)^2+(y-10)^2-0.01)(y-2x)=0 [-7.13, 12.87, 0.92, 10.92]}