Show What You Know

Question

Show What You Know
Analyze Media 
Carla is watching a professional golf game on television. A graphic overlay comes up during the replay that shows how high and far the ball traveled in feet.

Part A
Question 1

Analyze the graph to estimate how high the golf ball was above the ground at its highest point.

Question 2

Analyze the graph to estimate the how far away the ball landed from where it was hit.

Part B
Carla recognizes the shape of the golf ball's path to be a parabola, which is the graph of a quadratic equation. She remembers that these types of equations can be written in the standard form  y=ax2+bx+c
 .

Question 3

Use the table to explain to Carla how she can use the graph to determine if each coefficient in the ball’s motion equation is positive, negative, or zero.

Coefficient	Sign	Explanation
a
 
+−0

b
 
+−0

c
 
+−0

The quadratic equation that represents the path of the golf ball can be written as  y=−0.001958x2+0.8391x
 .

Question 4

How do the root(s) of this equation relate to the example of Carla’s golf ball scenario?

Question 5

Which method would you recommend that Carla use to solve this quadratic equation?

Question 6

Solve this quadratic equation using your recommended method.

GPT-4o mini · Answer

### Part A **Question 1:** To estimate how high the golf ball was above the ground at its highest point, you must check the graph overlay during the replay. Look for the maximum point on the graph where the vertical height (y-axis) is at its peak. Given that the ball's height would generally reach its maximum at the vertex of the parabola, identify the corresponding height value at that peak. **Answer:** The height of the golf ball at its highest point is the maximum value observed on the y-axis. (e.g., if the graph states 150 feet, then the highest point was 150 feet). --- **Question 2:** To estimate how far away the ball landed from where it was hit, find where the graph intersects the horizontal line (the ground level) on the x-axis. This point represents the horizontal distance the ball traveled from its starting position to where it landed. **Answer:** The distance from where the ball was hit to where it landed is the x-coordinate of the point where the graph meets the x-axis. (e.g., if the x-axis shows a distance of 400 feet, then the ball landed 400 feet away). --- ### Part B **Question 3:** To analyze the coefficients of the quadratic equation in the form $ y = ax^2 + bx + c $, you can determine the sign (positive, negative, or zero): - **Coefficient a:** If $ a $ is negative, the parabola opens downwards, indicating that the maximum point is at the vertex (which matches the golf ball's trajectory). If it were positive, it would open upwards, indicating a minimum point, which isn't what we see in a typical ball trajectory. - **Coefficient b:** The sign of $ b $ indicates the direction in which the parabola is leaning. A positive $ b $ suggests that the ball moves upwards initially before falling, while a negative $ b $ would indicate the opposite. - **Coefficient c:** This represents the y-intercept, or the height from which the ball was launched. If $ c $ is positive, the ball starts above ground, and if $ c $ is zero, it starts at ground level. | Coefficient | Sign | Explanation | |-------------|------|-------------| | a | - | The parabola opens downwards, confirming a peak height (max point). | | b | + | Indicates initial upward motion before falling. | | c | + | Starts from a height greater than zero (above ground). | --- **Question 4:** The roots of the equation $ y = -0.001958x^2 + 0.8391x $ correspond to the points where the golf ball hits the ground (y=0). These points indicate the distances along the horizontal axis where the ball left and reached the ground, showing where the ball was struck and where it landed. **Answer:** The roots represent how far the ball traveled before landing (where $ y=0 $) — the starting point and the landing point. --- **Question 5:** For solving this quadratic equation, the most straightforward method would be to use the quadratic formula: $$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} $$ This method will efficiently provide the exact roots of the equation even if they are irrational. **Answer:** I recommend using the quadratic formula to solve the equation. --- **Question 6:** Using the quadratic formula, plug in the values: $$ a = -0.001958, \quad b = 0.8391, \quad c = 0 $$ 1. Calculate the discriminant: $$ b^2 - 4ac = (0.8391)^2 - 4(-0.001958)(0) = 0.7035 $$ 2. Apply the formula: $$ x = \frac{{-0.8391 \pm \sqrt{0.7035}}}{2(-0.001958)} $$ 3. Calculate the roots: $$ x = \frac{{-0.8391 \pm 0.8391}}{-0.003916} $$ $$ x_1 = \frac{0}{-0.003916} = 0 \quad (\text{initial hit point, where the ball was hit}) $$ $$ x_2 = \frac{{-1.6782}}{-0.003916} \approx 428.2 \text{ feet} \quad (\text{where the ball lands}) $$ Therefore, the solved quadratic equation yields: **Answer:** The ball was hit at 0 feet and landed approximately at 428.2 feet away.

Answer

redo question 6

GPT-4o mini · Answer

Certainly! Let's solve the quadratic equation given as $ y = -0.001958x^2 + 0.8391x $ using the quadratic formula step by step.

Given Equation

The equation is:

\[ y = -0.001958x^2 + 0.8391x = 0 \]

Identify Coefficients

From the equation, we can identify the coefficients $ a $, $ b $, and $ c $:

$ a = -0.001958 $
$ b = 0.8391 $
$ c = 0 $

Quadratic Formula

The quadratic formula to find the roots of the equation $ ax^2 + bx + c = 0 $ is:

\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]

Step 1: Calculate the Discriminant

First, we calculate the discriminant $ D = b^2 - 4ac $:

\[ D = (0.8391)^2 - 4 \cdot (-0.001958) \cdot 0 \] \[ D = 0.7034 - 0 = 0.7034 \]

Step 2: Calculate the Roots

Now we can substitute the values into the quadratic formula:

\[ x = \frac{{-0.8391 \pm \sqrt{0.7034}}}{2 \cdot (-0.001958)} \]

First, calculate $ \sqrt{0.7034} $:

\[ \sqrt{0.7034} \approx 0.8392 \]

Now substitute this back into the formula:

\[ x = \frac{{-0.8391 \pm 0.8392}}{-0.003916} \]

Step 3: Calculate the Two Potential Roots

Using the positive root: \[ x_1 = \frac{{-0.8391 + 0.8392}}{-0.003916} \approx \frac{{0.0001}}{-0.003916} \approx -0.0255 \] (This root is not relevant for our scenario since time/distance cannot be negative.)
Using the negative root: \[ x_2 = \frac{{-0.8391 - 0.8392}}{-0.003916} \approx \frac{{-1.6783}}{-0.003916} \approx 428.4 \text{ feet} \]

Final Answer

Thus, the effective distances where the ball was hit (0 feet) and where it landed is approximately:

The ball was hit at $ 0 $ feet and landed at approximately $ 428.4 $ feet away.

Final Answer: The ball lands approximately 428.4 feet away from the initial hit point.