Question

note: you can interact with the applet by moving the purple \x\ on the horizontal axis to represent varying the value of x.
graph with (5,3), δx, δy, and purple x on x - axis
drag the purple x on the x - axis to vary the value of x
a. drag the purple x on the x - axis to vary the value of x from x = 5 to x = 3.
determine the value of δx as x varies from x = 5 to x = 3.
δx = -0.03 preview -0.03 = -0.03.
b. drag the purple x on the x - axis to vary the value of x from x = 5 to x = 3 and notice how y varies on this interval.
as the value x varies from x = 5 to x = 3, the value of y decreases from 0.06 to -2.
c. determine the value of δy as x varies from x = 5 to x = 3.
δy = blank preview
enter a mathematical expression more...
d. determine the value of \\(\frac{delta y}{delta x}\\) as x varies over the interval from x = 5 to x = 3.
\\(\frac{delta y}{delta x}\\) = blank preview

Explanation:

Part c: Determine $\Delta y$

Step1: Recall $\Delta y$ formula

$\Delta y = y_{\text{final}} - y_{\text{initial}}$

Step2: Identify $y$-values

At $x = 5$, $y = 3$ (from point $(5, 3)$). At $x = 3$, assume $y = 0$? Wait, no—wait, the purple X is at $x = 3$? Wait, no, the graph: when $x$ goes from 5 to 3, the $y$ at $x=5$ is 3, and at $x=3$, what's $y$? Wait, the red arrow: $\Delta y$ is change in $y$. Wait, maybe the point at $x=5$ is $(5, 3)$, and when $x$ is 3, what's $y$? Wait, maybe the vertical change: if $x$ goes from 5 to 3, and the $y$ at $x=5$ is 3, and at $x=3$, maybe $y=0$? No, wait, the purple X is on the x-axis, so $y=0$? Wait, no, the point $(5,3)$ is a point, and the purple X is on the x-axis (so $y=0$). So when $x$ is 5, the other point is $(5,3)$, and when $x$ is 3, the purple X is at $(3, 0)$? Wait, maybe: initial $y$ (at $x=5$) is 3, final $y$ (at $x=3$) is 0? Wait, no, the red arrow is $\Delta y$, which is final $y$ minus initial $y$. Wait, initial $x=5$, $y=3$; final $x=3$, $y=0$? Wait, no, maybe the purple X is at $x=3$, $y=0$, and the point is $(5,3)$. So $\Delta y = 0 - 3 = -3$? Wait, let's check:

Wait, the problem says "as $x$ varies from $x=5$ to $x=3$", so initial $x=5$, $y=3$; final $x=3$, what's $y$? The purple X is on the x-axis, so $y=0$ (since it's on the x-axis, $y$-coordinate is 0). So $\Delta y = y_{\text{final}} - y_{\text{initial}} = 0 - 3 = -3$.

Wait, but in part b, it says "from 0.06 to -2"—maybe I misread. Wait, no, maybe the point is $(5, 3)$, and when $x=3$, the $y$-value is, say, 0? Wait, maybe the user made a typo in part b, but let's proceed.

Wait, $\Delta y$ is change in $y$, so final $y$ minus initial $y$. Initial $x=5$, $y=3$; final $x=3$, $y=0$ (since purple X is on x-axis, $y=0$). So $\Delta y = 0 - 3 = -3$.

Step1: Recall $\Delta x$ from part a

From part a, $\Delta x = 3 - 5 = -2$? Wait, no—wait, $\Delta x = x_{\text{final}} - x_{\text{initial}} = 3 - 5 = -2$. Wait, in part a, the user wrote -0.03, but that's probably a typo. Wait, $x$ goes from 5 to 3, so $\Delta x = 3 - 5 = -2$.

Step2: Use $\Delta y$ from part c

$\Delta y = -3$ (from part c).

Step3: Calculate $\frac{\Delta y}{\Delta x}$

$\frac{\Delta y}{\Delta x} = \frac{-3}{-2} = \frac{3}{2} = 1.5$? Wait, no—wait, if $\Delta x = 3 - 5 = -2$, and $\Delta y = 0 - 3 = -3$, then $\frac{-3}{-2} = 1.5$. But wait, maybe the $\Delta x$ in part a was a typo. Let's recheck:

Wait, $x$ from 5 to 3: $\Delta x = 3 - 5 = -2$. So if $\Delta x = -2$, and $\Delta y = -3$, then $\frac{\Delta y}{\Delta x} = \frac{-3}{-2} = 1.5$.

But in part a, the user entered -0.03, which is incorrect. Probably a mistake. Let's assume correct $\Delta x = 3 - 5 = -2$, $\Delta y = 0 - 3 = -3$. Then $\frac{\Delta y}{\Delta x} = \frac{-3}{-2} = 1.5$.

Answer:

(Part c):
$\Delta y = \boxed{-3}$

Part d: Determine $\frac{\Delta y}{\Delta x}$