# How do you find the equation of the secant line of f(x)=x^2-5x through the points [1,8]?

Dec 20, 2016

The line will pass through the points $\left(1 , f \left(1\right)\right)$ and $\left(8 , f \left(8\right)\right)$

$f \left(1\right) = {1}^{2} - 5 \left(1\right)$

$f \left(1\right) = - 4$

$f \left(8\right) = {8}^{2} - 5 \left(8\right)$

$f \left(8\right) = 24$

The line will pass through the points $\left(1 , - 4\right)$ and $\left(8 , 24\right)$

The slope is $m = \frac{24 - \left(- 4\right)}{8 - 2} = \frac{28}{7} = 4$

The point-slope form of the equation of a line is:

$y = m \left(x - {x}_{0}\right) + {y}_{0}$

Substitute $4$ for m, 1 for ${x}_{0}$ and -4 for ${y}_{0}$

$y = 4 \left(x - 1\right) - 4$

$y = 4 x - 4 - 4$

$y = 4 x - 8$

Dec 20, 2016

$y = 4 x - 8$

#### Explanation:

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

When $x = 1 \implies f \left(x\right) = 1 - 5 = - 4$
When $x = 8 \implies f \left(x\right) = 64 - 40 = 24$

So the required secant line passes through the points $\left(1 , - 4\right)$ and $\left(8 , 24\right)$.

We can calculate the slope of the secant line using

$m = \frac{\Delta y}{\Delta x} = \frac{24 - \left(- 4\right)}{8 - 1} = \frac{28}{7} = 4$

So using the factthat the line passes through $\left(1 , - 4\right)$ and has slope $4$ and Using the formula $y - {y}_{0} = m \left(x - {x}_{0}\right)$, the required equation is given by:

$y - \left(- 4\right) = 4 \left(x - 1\right)$
$\therefore y + 4 = 4 x - 4$
$\therefore y = 4 x - 8$

NB: We could have equally used the other coordinate

Which we can confirm graphically: 