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

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Answer 1 · 2018-05-13T16:29:49

Slope of secant is $2$ $2$ .

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

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

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

Now slope of line joining $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$ is

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $(y_2-y_1)/(x_2-x_1)$

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

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]}

How do you find the slope of the secant lines of f(x)=x2−4x+5f (x) = x ^2 - 4x + 5 at (1, 2) and (5, 10)?