Question

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

Answer 1

The line will pass through the points `(1, f(1))` and `(8, f(8))`

`f(1) = 1^2 - 5(1)`

`f(1) = -4`

`f(8) = 8^2 - 5(8)`

`f(8) = 24`

The line will pass through the points `(1, -4)` and `(8, 24)`

The slope is `m =(24 - (-4))/(8 - 2) = 28/7 = 4`

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

`y = m(x - x_0) + y_0`

Substitute `4` for m, 1 for `x_0` and -4 for `y_0`

`y = 4(x - 1) - 4`

`y = 4x - 4 - 4`

`y = 4x - 8`

Answer 2

`y = 4x-8`

Explanation:

We have `f(x) = x^2 - 5x`

When `x=1 => f(x) = 1-5 = -4`
When `x=8 => f(x) = 64-40 = 24`

So the required secant line passes through the points `(1, -4)` and `(8, 24)`.

We can calculate the slope of the secant line using

`m=(Delta y)/(Delta x) = (24-(-4)) / (8-1) = 28/7 = 4`

So using the factthat the line passes through `(1,-4)` and has slope `4` and Using the formula `y - y_0 = m(x - x_0)`, the required equation is given by:

`y - (-4)=4(x-1)`
`:. y +4 = 4x-4`
`:. y = 4x-8`

NB: We could have equally used the other coordinate

Which we can confirm graphically: