We consider the function $y(x) = 5x^2 - 3x + 10$. The difference quotient is equal to 12 and $x=4$ is the starting point. Determine $\Delta x$.
We start in the same way as in Example 1, by determining $y(x) = y(4)$ and $y(x+\Delta x) = y(4+\Delta x)$:
$$\begin{align}
y(4) &= 5\cdot4^2 - 3\cdot4 + 10 = 78,\\
y(4+\Delta x) &= 5(4+\Delta x)^2 - 3(4+\Delta x) + 10 = 5(16 + 8\Delta x + (\Delta x)^2) - 3(4+\Delta x) + 10\\
&= 80 + 40\Delta x + 5(\Delta x)^2 - 12 - 3\Delta x + 10 = 5(\Delta x)^2 + 37\Delta x + 78.
\end{align}$$
Subsequently, we denote the difference quotient and put it equal to 12:
$$\begin{align}
\dfrac{\Delta y}{\Delta x} = \dfrac{y(4+\Delta x)-y(4)}{\Delta x} = \dfrac{5(\Delta x)^2 + 37\Delta x + 78 - 78}{\Delta x} = \dfrac{5(\Delta x)^2 + 37\Delta x}{\Delta x} &= 12\\
5\Delta x + 37 &= 12\\
5\Delta x &= -25\\
\Delta x &= -5.
\end{align}$$
Hence, $\Delta x=-5$; $x$ decreases by 5.
