Question

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)?

Answer 1

Find the slope of the line through the two points.

Explanation:

Slope of a line through points `(x_1,y_1)` and `(x_2,y_2)` is

`m = (y_2-y_1)/(x_2-x_1) = (Deltay)/(Deltax)`

We will need `f (x) = 2x^2 - 3x - 5`
to find the `y` values of the points: `(2, f (2))` and `(2 + h, f (2 + h))`

`m = (f(2+h)-f(2))/((2+h)-(2))`

`= (overbrace([2(2+h)^2-3(2+h)-5])^f(2+h) - overbrace([2(2)^2-3(2)-5])^f(2))/h`

`= ([2(4+4h+h^2)-3(2+h)-5] - [2(4)-3(2)-5])/h`

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

`= (8+8h+2h^2-6-3h-5-8+6+5)/h`

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

`= (color(red)(8)+8h+2h^2color(green)(-6)-3hcolor(blue)(-5)color(red)(-8)color(green)(+6)color(blue)(+5))/h`

`= (8h+2h^2-3h)/h`

`= (2h^2+5h)/h = (h(2h+5))/h`

`= 2h+5` `" "` for `h != 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.