A) Graph By First Finding The Vertex, Zero(s), Y Intercept Algebraically For F(x)=x-5x-6 B) A Diver Dives

A) Graph By First Finding The Vertex, Zero(s), Y Intercept Algebraically For F(x)=x-5x-6 B) A Diver Dives

Understanding how to graph quadratic functions and interpret real-world scenarios such as a diver diving involves a combination of algebraic techniques and practical analysis. In this comprehensive guide, we will explore how to graph the quadratic function \( F(x) = x - 5x - 6 \) by first finding key features like the vertex, zeroes, and y-intercept algebraically. Additionally, we will delve into the physics of a diver diving, analyzing the motion and trajectory, which often involves quadratic functions. Mastering these concepts enables students and enthusiasts to effectively interpret mathematical models and apply them to real-life situations.

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Part 1: Graphing the Quadratic Function Algebraically

Understanding the Function: Simplify and Clarify the Expression

The given quadratic function is:

\[ F(x) = x - 5x - 6 \]

Before proceeding with graphing, simplify the function:

\[ F(x) = (x - 5x) - 6 = -4x - 6 \]

This simplifies to a linear function, not quadratic. However, assuming the original intended function was quadratic, perhaps a typo occurred, and it was meant to be:

\[ F(x) = x^2 - 5x - 6 \]

Assuming the quadratic form, we'll proceed with:

\[ F(x) = x^2 - 5x - 6 \]

If the original was linear, the same steps apply for linear functions, but here, the focus is on quadratics.

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Step 1: Find the Vertex of the Quadratic Function

Using Vertex Formula

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using:

\[ x_{v} = -\frac{b}{2a} \]

For \( F(x) = x^2 - 5x - 6 \):

• \( a = 1 \)


• \( b = -5 \)

Calculate the x-coordinate of the vertex:

\[ x_{v} = -\frac{-5}{2 \times 1} = \frac{5}{2} = 2.5 \]

Now, find the y-coordinate by substituting \( x = 2.5 \) into \( F(x) \):

\[ F(2.5) = (2.5)^2 - 5(2.5) - 6 = 6.25 - 12.5 - 6 = -12.25 \]

Vertex Coordinates:

\[ (2.5, -12.25) \]

This point is the maximum or minimum of the parabola depending on the sign of \( a \). Since \( a = 1 > 0 \), the vertex is a minimum point.

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Step 2: Find the Zero(s) (X-Intercepts)

Zeros are the solutions to \( F(x) = 0 \):

\[ x^2 - 5x - 6 = 0 \]

Use the quadratic formula:

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

Plugging in the values:

\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times (-6)}}{2 \times 1} \]
\[ x = \frac{5 \pm \sqrt{25 + 24}}{2} \]
\[ x = \frac{5 \pm \sqrt{49}}{2} \]
\[ x = \frac{5 \pm 7}{2} \]

Calculate both solutions:

1. \( x = \frac{5 + 7}{2} = \frac{12}{2} = 6 \)


2. \( x = \frac{5 - 7}{2} = \frac{-2}{2} = -1 \)

Zeroes:

\[ x = -1, 6 \]

These points are where the parabola crosses the x-axis.

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Step 3: Find the Y-Intercept

The y-intercept occurs when \( x = 0 \):

\[ F(0) = (0)^2 - 5(0) - 6 = -6 \]

Y-Intercept:

\[ (0, -6) \]

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Part 2: Graphing the Quadratic Function

Step-by-Step Graphing Process

To sketch the graph of \( F(x) = x^2 - 5x - 6 \):

1. Plot the vertex at \( (2.5, -12.25) \).
2. Mark the zeroes at \( (-1, 0) \) and \( (6, 0) \).
3. Plot the y-intercept at \( (0, -6) \).
4. Draw the parabola passing through these points, ensuring symmetry about the axis of symmetry at \( x = 2.5 \).

Additional Points for Accuracy

To improve the graph's accuracy, find additional points:

• At \( x = 1 \):

\[ F(1) = 1 - 5 - 6 = -10 \]

• At \( x = 4 \):

\[ F(4) = 16 - 20 - 6 = -10 \]

These symmetric points confirm the parabola's shape.

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Part 3: Application — A Diver Dives

Modeling a Diver’s Dive with Quadratic Functions

In physics, the motion of a diver diving off a platform or cliff can be modeled using quadratic functions, especially when analyzing the vertical component of the motion under gravity.

Key Concepts:

• The height of the diver at time \( t \) can be modeled as:

\[ h(t) = -\frac{1}{2} g t^2 + v0 t + h0 \]

where:

• \( g \) is acceleration due to gravity (\( \approx 9.8 \, m/s^2 \))


• \( v_0 \) is the initial velocity (upward or downward)


• \( h_0 \) is the initial height

Scenario:

Suppose a diver jumps downward from a platform, and their height \( h(t) \) in meters after \( t \) seconds is modeled by:

\[ h(t) = -4.9 t^2 + v0 t + h0 \]

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Analyzing the Diver’s Motion

1. Initial Height \( h_0 \): height of the platform
2. Initial Velocity \( v_0 \): speed at which the diver leaves the platform
3. Time to Hit the Water: solve for \( t \) when \( h(t) = 0 \)
4. Maximum Height: occurs at the vertex of the parabola

Example:

If a diver jumps with an initial velocity \( v_0 = 2 \, m/s \) from a platform 10 meters high:

\[ h(t) = -4.9 t^2 + 2 t + 10 \]

• Find the time when the diver hits the water:

\[ 0 = -4.9 t^2 + 2 t + 10 \]

Apply quadratic formula:

\[ t = \frac{-2 \pm \sqrt{(2)^2 - 4 \times (-4.9) \times 10}}{2 \times -4.9} \]
\[ t = \frac{-2 \pm \sqrt{4 + 196}}{-9.8} \]
\[ t = \frac{-2 \pm \sqrt{200}}{-9.8} \]
\[ t = \frac{-2 \pm 14.14}{-9.8} \]

Calculate both:

• \( t = \frac{-2 + 14.14}{-9.8} = \frac{12.14}{-9.8} \approx -1.24 \) (discard negative time)


• \( t = \frac{-2 - 14.14}{-9.8} = \frac{-16.14}{-9.8} \approx 1.65 \)

Time to hit the water: approximately 1.65 seconds.

• Find maximum height:

Vertex occurs at:

\[ t{v} = -\frac{v0}{2a} = -\frac{2}{2 \times -4.9} = -\frac{2}{-9.8} \approx 0.204 \, s \]

At \( t = 0.204 \):

\[ h(0.204) = -4.9 (0.204)^2 + 2 (0.204) + 10 \]
\[ h(0.204) \approx -4.9 (0.0416) + 0.408 + 10 \]
\[ h(0.204) \approx -0.204 + 0.408 + 10 \]
\[ h(0.204) \approx 10.204 \, m \]

The diver reaches