a. what is the initial balance in the account, to the nearest hundred?
b. is the account modeled by a linear function? explain how you know.
c. a second investment account starts with $1,000 and grows by $25 each year. sketch the values of this account on the graph.
d. are the two account balances ever equal? what happens in the years after that? explain how you know.

Question

Accepted Answer

To solve these problems, we assume there is a graph (not shown here) representing the first account's balance over time. We'll address each sub - question: ### Part a # Explanation: ## Step 1: Analyze the graph's y - intercept The initial balance is the value of the account when time ($t$) = 0, which is the y - intercept of the graph representing the account's balance. We look at the y - axis (balance) value when $t = 0$ and round it to the nearest hundred. (Assuming from typical problems, if the y - intercept is around, say, $1000$ (but since the graph is not shown, we'll assume a common case. If the graph starts at a value like 950, it rounds to 1000; if it's 1040, it rounds to 1000, etc. For the sake of this example, we'll assume the initial balance from the graph (when $t = 0$) is approximately $1000$ when rounded to the nearest hundred. ) # Answer: The initial balance, to the nearest hundred, is $\$1000$ (this answer is based on the assumption of a typical graph for such problems. The actual value depends on the given graph). ### Part b # Explanation: ## Step 1: Recall the definition of a linear function A linear function has a constant rate of change (slope). Mathematically, a linear function can be written in the form $y=mx + b$, where $m$ (the slope) is constant and $b$ is the y - intercept. ## Step 2: Check the rate of change of the account balance To determine if the account is modeled by a linear function, we check if the change in balance ( $\Delta y$) per unit change in time ( $\Delta t$) is constant. If the graph of the account's balance over time is a straight line, then the rate of change (slope) is constant, and the function is linear. If the graph is curved, the rate of change is not constant, and it is not a linear function. (For example, if the balance increases by the same amount each year, the graph is a straight line, and it is linear. If the balance increases by a different amount each year (e.g., exponential growth where the amount of increase is a percentage of the current balance), the graph is curved, and it is not linear.) # Answer: To determine if the account is modeled by a linear function, we check the rate of change of the balance with respect to time. If the graph of the account's balance over time is a straight line (indicating a constant rate of change of balance per unit time), then the account is modeled by a linear function. If the graph is curved (indicating a non - constant rate of change), then it is not. (The final determination depends on the shape of the graph of the account's balance over time.) ### Part c # Explanation: ## Step 1: Determine the equation of the second account The second account starts with $b_0=\$1000$ and has a rate of change (slope) $m = \$25$ per year. The equation of a linear function is $y=mx + b$, where $y$ is the balance, $x$ is the time in years, $m$ is the rate of change, and $b$ is the initial balance. So the equation for the second account is $y = 25x+1000$. ## Step 2: Plot points for the second account - When $x = 0$ (initial time), $y=25(0)+1000 = 1000$. So the point is $(0,1000)$. - When $x = 1$, $y=25(1)+1000=1025$. The point is $(1,1025)$. - When $x = 2$, $y=25(2)+1000 = 1050$. The point is $(2,1050)$. We then sketch a straight line passing through these points on the same graph as the first account. # Answer: To sketch the second account: 1. Plot the point $(0,1000)$ (since at $t = 0$, the balance is $\$1000$). 2. Then, for each subsequent year ($t = 1,2,3,\cdots$), move up $\$25$ (since the balance grows by $\$25$ per year) from the previous point and plot the new point. For example, at $t = 1$, plot $(1,1025)$; at $t = 2$, plot $(2,1050)$, etc. Then draw a straight line through these points. ### Part d # Explanation: ## Step 1: Analyze the two functions Let the first account's balance be $y_1$ and the second account's balance be $y_2=25t + 1000$ (where $t$ is time in years). The first account's function is $y_1$ (from its graph). ## Step 2: Check for intersection To see if the two account balances are ever equal, we need to check if the graphs of $y_1$ and $y_2$ intersect. If the first account is linear (from part b, if it is linear) with a slope $m_1$: - If $m_1=m_2 = 25$, and $b_1=b_2$, the lines are coincident and all points are equal. But if $m_1 eq25$: - If $m_1>25$: The first account grows faster. If at $t = 0$, $y_1(0)=y_2(0) = 1000$ (from part a and the second account's initial balance), and $m_1>25$, then $y_1$ will be greater than $y_2$ for all $t>0$. - If $m_1<25$: The second account grows faster. If at $t = 0$, $y_1(0)=y_2(0)=1000$, and $m_1 < 25$, then $y_2$ will be greater than $y_1$ for all $t>0$. - If the first account is non - linear (e.g., exponential), say $y_1 = a(1 + r)^t+b$, we can set $a(1 + r)^t+b=25t + 1000$ and solve for $t$. But in most cases, if the first account is linear with a different slope: (Assuming the first account is linear with a slope different from 25. For example, if the first account has a slope of 30 (grows by $30$ per year) and initial balance $1000$, then $y_1=30t + 1000$ and $y_2=25t + 1000$. Setting $30t+1000 = 25t + 1000$ gives $5t=0$ or $t = 0$. So they are equal only at $t = 0$. If the first account has a slope of 20 (grows by $20$ per year), $y_1=20t + 1000$ and $y_2=25t + 1000$. Setting $20t + 1000=25t+1000$ gives $- 5t = 0$ or $t = 0$. But if we solve $20t+1000=25t + 1000$ for $t>0$, we get no solution. However, if we consider non - linear first account, say $y_1=1000(1.02)^t$ (exponential growth with 2% interest) and $y_2=25t + 1000$. We can find $t$ such that $1000(1.02)^t=25t + 1000$. By testing values: At $t = 0$, both are 1000. At $t = 1$, $y_1=1020$, $y_2 = 1025$ (so $y_2>y_1$). At $t = 10$, $y_1=1000(1.02)^{10}\approx1219.0$, $y_2=25 imes10 + 1000=1250$ ($y_2>y_1$). At $t = 20$, $y_1=1000(1.02)^{20}\approx1485.9$, $y_2=25 imes20 + 1000 = 1500$ ($y_2>y_1$). At $t = 30$, $y_1=1000(1.02)^{30}\approx1811.4$, $y_2=25 imes30+1000 = 1750$ ($y_1>y_2$). So there will be a time $t$ between 20 and 30 when they are equal. But in the case of two linear functions with the same initial balance: If the slopes are different: - If $m_1>m_2$: The first account will be greater than the second for $t>0$. - If $m_10$. - If $m_1 = m_2$: The two accounts will have the same balance for all $t$. (Assuming the first account is linear with a slope different from 25 and same initial balance of 1000. Let's say the first account has a slope of $m_1 eq25$) # Answer: - If the two accounts are both linear with the same initial balance: - If their slopes (rates of change of balance per year) are equal, the two account balances will be equal for all time $t$. - If their slopes are different: - If the slope of the first account is greater than 25 (the slope of the second account), the first account's balance will be greater than the second account's balance for all $t>0$, and they are only equal at $t = 0$. - If the slope of the first account is less than 25, the second account's balance will be greater than the first account's balance for all $t>0$, and they are only equal at $t = 0$. - If the first account is non - linear (e.g., exponential), there may be a time $t>0$ when the two account balances are equal. After that time, the account with the faster growth rate (either the first or the second, depending on the functions) will have a greater balance.

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QUESTION IMAGE

Question

Explanation:

Part a

Step 1: Analyze the graph's y - intercept

Step 1: Recall the definition of a linear function

Step 2: Check the rate of change of the account balance

Step 1: Determine the equation of the second account

Step 2: Plot points for the second account

Answer:

Part b