...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as: $$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$ (the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper} = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\ \text{and }\\ \mathrm{rate}_{lower} = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1} = \pu{2 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}\\ \text{and }\\ \mathrm{rate}_{2} = \pu{4 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}$$

Since a reaction of $\mathrm{rate}_1$ has a more negative slope, it must correspond to a curve that decreases faster than a reaction of $\mathrm{rate}_2$ and the concentration would be lower sooner, thus $\mathrm{rate}_1$ is the lower curve and $\mathrm{rate}_2$ is the upper curve.

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as: $$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$ (the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper} = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\ \text{and }\\ \mathrm{rate}_{lower} = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1} = \pu{2 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}\\ \text{and }\\ \mathrm{rate}_{2} = \pu{4 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}$$

Since a reaction of $\mathrm{rate}_2$ has a more negative slope, it must correspond to a curve that decreases faster than a reaction of $\mathrm{rate}_1$ and the concentration would be lower sooner, thus $\mathrm{rate}_2$ is the lower curve and $\mathrm{rate}_1$ is the upper curve.

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as: $$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$ (the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as: $$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$ (the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper} = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\ \text{and }\\ \mathrm{rate}_{lower} = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1} = \pu{2 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}\\ \text{and }\\ \mathrm{rate}_{2} = \pu{4 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}$$

Since a reaction of $\mathrm{rate}_1$ has a more negative slope, it must correspond to a curve that decreases faster than a reaction of $\mathrm{rate}_2$ and the concentration would be lower sooner, thus $\mathrm{rate}_1$ is the lower curve and $\mathrm{rate}_2$ is the upper curve.