Regression Key | PDF - Free Printable

Here are the step-by-step solutions and checks for the problems on the worksheet.

Problem 1 & 2: Basketball Trajectory

Step 1: Find the quadratic equation.
We need to find an equation in the form $f(x) = ax^2 + bx + c$ that fits the data points provided in the table (Distance vs. Height).
Using a calculator or regression tool with the points $(0, 4), (2, 8.4), (6, 12.1), (9, 14.2), (12, 13.2), (13, 10.5), (15, 9.8)$:

* The y-intercept ($c$) is clearly 4 because when $x=0$, $y=4$.
* Running the quadratic regression yields coefficients very close to:
* $a \approx -0.118$
* $b \approx 2.112$
* $c \approx 4.215$ (Note: While the table shows 4 at x=0, regression finds the "best fit" line which might adjust the intercept slightly to minimize error across all points. The handwritten key uses 4.215).

So, the equation is:
$$f(x) = -0.118x^2 + 2.112x + 4.215$$

Step 2: Find the maximum height.
The maximum height of a parabola occurs at the vertex. The x-coordinate of the vertex is found using $x = \frac{-b}{2a}$.

* $a = -0.118$
* $b = 2.112$

$$x = \frac{-2.112}{2(-0.118)} = \frac{-2.112}{-0.236} \approx 8.949$$

Now, plug this x-value back into the equation to find the height ($y$):
$$f(8.949) = -0.118(8.949)^2 + 2.112(8.949) + 4.215$$
$$f(8.949) = -0.118(80.085) + 18.900 + 4.215$$
$$f(8.949) = -9.450 + 18.900 + 4.215$$
$$f(8.949) \approx 13.665$$

The approximate maximum height is 13.665 feet.

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Problem 3 & 4: City Population

Step 1: Find the best-fitting quadratic model.
We use the data points where $x$ is years since 1970 and $y$ is population in thousands:
$(0, 489), (10, 801), (20, 1202), (30, 1998), (40, 2959)$.

Running quadratic regression on these points:
* $a \approx 1.209$
* $b \approx 13.000$
* $c \approx 504.257$

The model is:
$$y = 1.209x^2 + 13.000x + 504.257$$

Step 2: Estimate the population in 2020.
First, determine the value of $x$ for the year 2020.
$$x = 2020 - 1970 = 50$$

Now, substitute $x = 50$ into the equation:
$$y = 1.209(50)^2 + 13.000(50) + 504.257$$
$$y = 1.209(2500) + 650 + 504.257$$
$$y = 3022.5 + 650 + 504.257$$
$$y = 4176.757$$

Rounding to one decimal place as shown in the key: 4,176.8 thousand.

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Problem 5: Identifying Quadratic Relationships

We need to identify which relationship follows the form $y = ax^2 + bx + c$.

* A. Circumference and diameter: $C = \pi d$. This is linear ($y = mx$).
* B. Area of a square and side length: $A = s^2$. This is quadratic ($y = x^2$).
* C. Diagonal and side length: $d^2 = s^2 + s^2 \rightarrow d = s\sqrt{2}$. This is linear.
* D. Volume of a cube and side length: $V = s^3$. This is cubic.

The correct choice is B.

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Problem 6: Completing the Table

We are given a table for a quadratic function and need to find the missing $y$ value when $x = 6$.
Points given: $(-2, -8), (0, 0), (2, 12), (4, 28)$.

Let's find the equation $y = ax^2 + bx + c$.
1. Since $(0,0)$ is a point, $c = 0$. So, $y = ax^2 + bx$.
2. Use point $(2, 12)$:
$$12 = a(2)^2 + b(2) \rightarrow 12 = 4a + 2b \rightarrow 6 = 2a + b$$
3. Use point $(4, 28)$:
$$28 = a(4)^2 + b(4) \rightarrow 28 = 16a + 4b \rightarrow 7 = 4a + b$$

Now solve the system:
$$b = 6 - 2a$$
Substitute into the second equation:
$$7 = 4a + (6 - 2a)$$
$$7 = 2a + 6$$
$$1 = 2a \rightarrow a = 0.5$$

Find $b$:
$$b = 6 - 2(0.5) = 6 - 1 = 5$$

The equation is $y = 0.5x^2 + 5x$.

Check with point $(-2, -8)$:
$$y = 0.5(-2)^2 + 5(-2) = 0.5(4) - 10 = 2 - 10 = -8$$ (Correct).

Now, calculate $y$ for $x = 6$:
$$y = 0.5(6)^2 + 5(6)$$
$$y = 0.5(36) + 30$$
$$y = 18 + 30$$
$$y = 48$$

The missing value is 48.

Final Answer:
1. $f(x) = -0.118x^2 + 2.112x + 4.215$
2. 13.665 feet
3. $y = 1.209x^2 + 13.000x + 504.257$
4. 4,176.8 thousand
5. B
6. D (48)