Question

problem: when planted, 15 inches long; two weeks later it was 22 inches, four weeks it was 36 inches graph data + write equation to represent

Explanation:

Step1: Identify variables and data points

Let \( x \) be the number of weeks, \( y \) be the length in inches.
Data points:

• When planted (\( x = 0 \)): \( y = 15 \)
• Two weeks (\( x = 2 \)): \( y = 22 \)
• Four weeks (\( x = 4 \)): \( y = 36 \)

Wait, check the rate. From \( x=0 \) to \( x=2 \): \( \frac{22 - 15}{2 - 0} = \frac{7}{2} = 3.5 \)
From \( x=2 \) to \( x=4 \): \( \frac{36 - 22}{4 - 2} = \frac{14}{2} = 7 \). Wait, maybe a typo? Or maybe linear? Wait, maybe the data is: planted (0 weeks):15, 2 weeks:22, 4 weeks:36? Wait, no, maybe 2 weeks later (x=2) y=22, 4 weeks later (x=4) y=36? Wait, let's re-express.

Wait, maybe the problem is: when planted (time \( t = 0 \)), length \( L = 15 \) inches. Two weeks later (\( t = 2 \)), \( L = 22 \) inches. Four weeks later (\( t = 4 \)), \( L = 36 \) inches. Wait, but the slope between (0,15) and (2,22) is \( m_1 = \frac{22 - 15}{2 - 0} = 3.5 \). Between (2,22) and (4,36), slope \( m_2 = \frac{36 - 22}{4 - 2} = 7 \). Hmm, not linear. Wait, maybe a mistake in the problem? Or maybe the third point is wrong? Wait, maybe it's 2 weeks:22, 4 weeks:36? Wait, 22 to 36 is 14 over 2 weeks, so 7 per week. From 15 to 22: 7 over 2 weeks, 3.5 per week. That's inconsistent. Wait, maybe the original problem has a typo, but let's assume it's linear. Wait, maybe the third point is 36? Wait, maybe the user made a typo. Alternatively, maybe the data is (0,15), (2,22), (4,36). Let's check if it's quadratic. Let's assume \( y = ax^2 + bx + c \). At \( x=0 \), \( c = 15 \). At \( x=2 \): \( 4a + 2b + 15 = 22 \Rightarrow 4a + 2b = 7 \). At \( x=4 \): \( 16a + 4b + 15 = 36 \Rightarrow 16a + 4b = 21 \). Solve:
From first equation: \( 2a + b = 3.5 \Rightarrow b = 3.5 - 2a \).
Substitute into second equation: \( 16a + 4(3.5 - 2a) = 21 \Rightarrow 16a + 14 - 8a = 21 \Rightarrow 8a = 7 \Rightarrow a = \frac{7}{8} = 0.875 \). Then \( b = 3.5 - 2(0.875) = 3.5 - 1.75 = 1.75 \). So equation: \( y = 0.875x^2 + 1.75x + 15 \). But maybe the problem intended linear? Wait, maybe the third point is 31? Because 15 + 3.52=22, 22 + 3.52=29? No. Wait, maybe the user wrote 36 instead of 29? Or 36 is correct. Alternatively, maybe the time is "two weeks later" (x=2) y=22, "four weeks later" (x=4) y=36. Let's proceed.

Step2: Graph the data points

Plot the points: (0,15), (2,22), (4,36) on a coordinate plane with x-axis as weeks, y-axis as length.

Step3: Find the equation (assuming quadratic, since linear slope changes)

We have \( y = ax^2 + bx + c \).
At \( x=0 \), \( y=15 \Rightarrow c = 15 \).
At \( x=2 \), \( y=22 \Rightarrow 4a + 2b + 15 = 22 \Rightarrow 4a + 2b = 7 \) (Equation 1).
At \( x=4 \), \( y=36 \Rightarrow 16a + 4b + 15 = 36 \Rightarrow 16a + 4b = 21 \) (Equation 2).

Multiply Equation 1 by 2: \( 8a + 4b = 14 \) (Equation 3).
Subtract Equation 3 from Equation 2: \( (16a + 4b) - (8a + 4b) = 21 - 14 \Rightarrow 8a = 7 \Rightarrow a = \frac{7}{8} = 0.875 \).

Substitute \( a = 0.875 \) into Equation 1: \( 4(0.875) + 2b = 7 \Rightarrow 3.5 + 2b = 7 \Rightarrow 2b = 3.5 \Rightarrow b = 1.75 \).

Thus, the equation is \( y = 0.875x^2 + 1.75x + 15 \), or in fraction form: \( y = \frac{7}{8}x^2 + \frac{7}{4}x + 15 \).

Wait, but maybe the problem intended linear, and there's a typo. Let's check: if linear, slope \( m = \frac{22 - 15}{2 - 0} = 3.5 \), so equation \( y = 3.5x + 15 \). Then at \( x=4 \), \( y = 3.5*4 + 15 = 14 + 15 = 29 \), but the problem says 36. So maybe the data is correct, and it's quadratic.

Alternatively, maybe the time is "after two weeks" (x=2) y=22, "after four weeks" (x=4) y=36. So the…

Answer:

To graph: Plot (0,15), (2,22), (4,36).
Equation: \( \boldsymbol{y = \frac{7}{8}x^2 + \frac{7}{4}x + 15} \) (or \( y = 0.875x^2 + 1.75x + 15 \))