What is the slope of the secant line of the function $y = −2x^2 + 3x − 1$ between $x=x_1$ and $x=x_2$?

Question

Ans:

1. $-2x_1-2x_2+3$
2. $-2x_1+2x_2+3$
3. $2x_1-2x_2+3$

I get that we can use the slope formula to get it to the state below. But how do we simplify it down to the final state (one of the answers)? Where is my knowledge gap on this one?

The slope of the secant line of the function $$y(x) = -2x^2 + 3x - 1$$ between points $x = x_1$ and $x=x_{2}$ is calculated as \begin{align*} m &= \frac{f(x_2) - f(x_1)}{x_2 - x_1}\\ &= \frac{-2x_2^2 + 3x^2 - 1 + 2x_1^2 - 3x_1 + 1}{x_2 - x_1}\\ &= \frac{3x_2 + 2x_1^2 - 3x_1 - 2x_2}{x_2 - x_1} \end{align*}

Answer 1

There is a typo:

\begin{align} \frac{3x_2+2x_1^2-3x_1-2x_2^{\color{red}2}}{x_2-x_1} &= \frac{3(x_2-x_1)-2(x_2-x_1)(x_2+x_1)}{x_2-x_1} \end{align}

Try to complete the task.

Answer 2

The given expression is

$\displaystyle\frac{-2x^2_2+3x_2-1+2x^2_1-3x_1+1}{x_2-x_1}$

$\displaystyle\frac{-2x^2_2+3x_2+2x^2_1-3x_1}{x_2-x_1}$

$\displaystyle\frac{-2x^2_2+3x_2+2x^2_1-3x_1}{x_2-x_1}$

$\displaystyle\frac{2(x_1-x_2)(x_1+x_2)-3(x_1-x_2)}{x_2-x_1}$

$\displaystyle\frac{(x_1-x_2)(2x_1+2x_2-3)}{x_2-x_1}$

$\displaystyle\frac{-(x_2-x_1)(2x_1+2x_2-3)}{x_2-x_1}$

$\displaystyle-2x_1-2x_2+3$