drag each equation to the correct location on the graph. not all equations will be used. match each equation…

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=0.1x^{3}+2$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=0.3x^{2}$. This gives the slope of the tangent line at any point $x$ on the curve.

Step2: Analyze the slopes of the given tangent - line equations

The equations of the tangent lines are in the form $y = mx + b$, where $m$ is the slope. For example, for $y = 0.3x+18$, the slope $m = 0.3$; for $y = 2.7x + 7.4$, the slope $m = 2.7$; for $y = 12x+3.6$, the slope $m = 12$; for $y = 4.8x - 108$, the slope $m = 4.8$.

Step3: Find the $x$ - values corresponding to the slopes

Set $f'(x)=0.3x^{2}$ equal to the slopes of the tangent - line equations. If $0.3x^{2}=0.3$, then $x^{2}=1$, so $x=\pm1$. If $0.3x^{2}=2.7$, then $x^{2}=9$, so $x = \pm3$. If $0.3x^{2}=12$, then $x^{2}=40$, so $x=\pm\sqrt{40}\approx\pm6.32$. If $0.3x^{2}=4.8$, then $x^{2}=16$, so $x=\pm4$.

Step4: Find the $y$ - values on the original function

For $x = 1$, $f(1)=0.1\times1^{3}+2=2.1$. For $x=-1$, $f(-1)=0.1\times(-1)^{3}+2 = 1.9$. For $x = 3$, $f(3)=0.1\times3^{3}+2=0.1\times27 + 2=4.7$. For $x=-3$, $f(-3)=0.1\times(-3)^{3}+2=-2.7 + 2=-0.7$. For $x = 4$, $f(4)=0.1\times4^{3}+2=0.1\times64 + 2=8.4$. For $x=-4$, $f(-4)=0.1\times(-4)^{3}+2=-6.4 + 2=-4.4$.

Step5: Use the point - slope form $y - y_0=m(x - x_0)$ to check the equations

For example, if $x = 3$, $m = 2.7$, and $y_0 = 4.7$, then $y-4.7=2.7(x - 3)$, $y-4.7=2.7x-8.1$, $y=2.7x - 3.4$. This is not one of the given equations. If $x=-3$, $m = 2.7$, and $y_0=-0.7$, then $y+0.7=2.7(x + 3)$, $y+0.7=2.7x+8.1$, $y=2.7x + 7.4$. If $x = 4$, $m = 4.8$, and $y_0 = 8.4$, then $y - 8.4=4.8(x - 4)$, $y-8.4=4.8x-19.2$, $y=4.8x - 10.8$. If $x=-4$, $m = 4.8$, and $y_0=-4.4$, then $y + 4.4=4.8(x + 4)$, $y+4.4=4.8x+19.2$, $y=4.8x + 14.8$ (not one of the given equations). If $x = 1$, $m = 0.3$, and $y_0 = 2.1$, then $y - 2.1=0.3(x - 1)$, $y-2.1=0.3x-0.3$, $y=0.3x + 1.8$ (not one of the given equations). If $x=-1$, $m = 0.3$, and $y_0 = 1.9$, then $y - 1.9=0.3(x + 1)$, $y-1.9=0.3x+0.3$, $y=0.3x+2.2$ (not one of the given equations). We assume we can also use the graph to estimate the slopes and intercepts visually. By visual inspection and calculation:

• The tangent line with a relatively small positive slope around $x=-3$ is $y = 2.7x+7.4$.
• The tangent line with a larger positive slope around $x = 4$ is $y = 4.8x - 108$.

Answer:

Match the equations to the graph based on slope and visual inspection as described above. Without the ability to actually drag the equations, the key is to note that for $y = 2.7x+7.4$, the slope is $2.7$ and it likely corresponds to a tangent point around $x=-3$ on the graph of $y = 0.1x^{3}+2$, and for $y = 4.8x - 108$, the slope is $4.8$ and it likely corresponds to a tangent point around $x = 4$ on the graph of $y = 0.1x^{3}+2$.