Check if you are correct by writing an equation or function that models the data in terms of "x"

Question

. the graph shows a line with a point of (0, 500) and a point of (1, 450) and a point at (2, 405)

Answer 1

To find a linear equation that models the data based on the given points, we can start by identifying the two points and using the slope-intercept form of the equation of a line \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

Given points:

1. \( (0, 500) \)
2. \( (1, 450) \)
3. \( (2, 405) \)

Step 1: Calculate the slope \( m \)

We can use the first two points \( (0, 500) \) and \( (1, 450) \) to calculate the slope:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{450 - 500}{1 - 0} = \frac{-50}{1} = -50 \]

Step 2: Determine y-intercept \( b \)

Since we know the line passes through the point \( (0, 500) \), the y-intercept \( b \) is simply 500.

Step 3: Write the equation

Using the slope \( m = -50 \) and y-intercept \( b = 500 \), we can write the equation:

\[ y = -50x + 500 \]

Step 4: Verify with other points

Now, let's check if this equation fits the other point \( (2, 405) \).

Substituting \( x = 2 \):

\[ y = -50(2) + 500 = -100 + 500 = 400 \]

However, this does not match the point \( (2, 405) \). It indicates that the line could be nonlinear.

Step 5: Try quadratic fitting

Using points \( (0, 500) \), \( (1, 450) \), and \( (2, 405) \), we can find a quadratic function of the form:

\[ y = ax^2 + bx + c \]

Since we know \( c \) (the value when \( x = 0 \)) is 500, we rewrite:

\[ y = ax^2 + bx + 500 \]

Now we need to set up a system of equations using the points \( (1, 450) \) and \( (2, 405) \):

1. For \( (1, 450) \): \[ 450 = a(1^2) + b(1) + 500 \Rightarrow a + b + 500 = 450 \Rightarrow a + b = -50 \quad (1) \]

2. For \( (2, 405) \): \[ 405 = a(2^2) + b(2) + 500 \Rightarrow 4a + 2b + 500 = 405 \Rightarrow 4a + 2b = -95 \quad (2) \]

Step 6: Solve the equations

From equation (1): \[ b = -50 - a \quad (3) \]

Substituting (3) into (2):

\[ 4a + 2(-50 - a) = -95 \ 4a - 100 - 2a = -95 \ 2a - 100 = -95 \ 2a = 5 \ a = 2.5 \]

Now substitute \( a \) back into (3):

\[ b = -50 - 2.5 = -52.5 \]

Final Model Equation

Thus, the quadratic equation that best fits the points is:

\[ y = 2.5x^2 - 52.5x + 500 \]

This quadratic function models the descent from the point (500) at \( x=0 \) to 450 at \( x=1 \) and 405 at \( x=2 \).