Draw A Diagram Of The Archway Modeled By The Equation Y = − X 2 + 5 X + 24 Y = -x^2 + 5x + 24 Y = − X 2 + 5 X + 24 .1. Find And Label The Y-intercept And The X-intercepts On The Sketch.2. Find And Label The Width Of The Archway At Its Base.3. Find And Label The Height Of The Archway

Drawing a Diagram of the Archway Modeled by the Equation $y = -x^2 + 5x + 24$

Understanding the Equation

The given equation $y = -x^2 + 5x + 24$ represents a quadratic function in the form of $y = ax^2 + bx + c$, where $a = -1$, $b = 5$, and $c = 24$. This equation models the shape of an archway, and our goal is to draw a diagram of this archway, labeling its key features.

Finding the y-Intercept

The y-intercept is the point where the archway intersects the y-axis. To find the y-intercept, we set $x = 0$ in the equation $y = -x^2 + 5x + 24$. This gives us:

$$y = -(0)^2 + 5(0) + 24 = 24$$

So, the y-intercept is at the point $(0, 24)$.

Finding the x-Intercepts

The x-intercepts are the points where the archway intersects the x-axis. To find the x-intercepts, we set $y = 0$ in the equation $y = -x^2 + 5x + 24$. This gives us a quadratic equation:

$$-x^2 + 5x + 24 = 0$$

We can solve this equation using the quadratic formula:

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

Plugging in the values $a = -1$, $b = 5$, and $c = 24$, we get:

$$x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(24)}}{2(-1)}$$

Simplifying, we get:

$$x = \frac{-5 \pm \sqrt{25 + 96}}{-2}$$

$$x = \frac{-5 \pm \sqrt{121}}{-2}$$

$$x = \frac{-5 \pm 11}{-2}$$

This gives us two possible values for $x$:

$$x = \frac{-5 + 11}{-2} = \frac{6}{-2} = -3$$

$$x = \frac{-5 - 11}{-2} = \frac{-16}{-2} = 8$$

So, the x-intercepts are at the points $(-3, 0)$ and $(8, 0)$.

Finding the Width of the Archway at its Base

The width of the archway at its base is the distance between the x-intercepts. Since the x-intercepts are at $x = -3$ and $x = 8$, the width of the archway at its base is:

$$\text{Width} = 8 - (-3) = 8 + 3 = 11$$

So, the width of the archway at its base is $11$ units.

Finding the Height of the Archway

The height of the archway is the maximum value of the function $y = -x^2 + 5x + 24$. To find the height, we need to find the vertex of the parabola. The x-coordinate of the vertex is given by:

$$x = \frac{-b}{2a} = \frac{-5}{2(-1)} = \frac{-5}{-2} = \frac{5}{2}$$

Plugging this value into the equation $y = -x^2 + 5x + 24$, we get:

$$y = -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) + 24$$

Simplifying, we get:

$$y = -\frac{25}{4} + \frac{25}{2} + 24$$

$$y = -\frac{25}{4} + \frac{50}{4} + \frac{96}{4}$$

$$y = \frac{121}{4}$$

So, the height of the archway is $\frac{121}{4}$ units.

Drawing the Diagram

Now that we have found the key features of the archway, we can draw a diagram of the archway. The diagram should include the following features:

• The y-intercept at $(0, 24)$
• The x-intercepts at $(-3, 0)$ and $(8, 0)$
• The width of the archway at its base, which is $11$ units
• The height of the archway, which is $\frac{121}{4}$ units

The diagram should also include the equation $y = -x^2 + 5x + 24$ to represent the shape of the archway.

Conclusion

In this article, we have drawn a diagram of the archway modeled by the equation $y = -x^2 + 5x + 24$. We have found and labeled the y-intercept and the x-intercepts on the sketch, and we have found and labeled the width of the archway at its base and the height of the archway. The diagram provides a visual representation of the shape of the archway and its key features.
Q&A: Drawing a Diagram of the Archway Modeled by the Equation $y = -x^2 + 5x + 24$

Frequently Asked Questions

Q: What is the equation of the archway? A: The equation of the archway is $y = -x^2 + 5x + 24$.

Q: What is the y-intercept of the archway? A: The y-intercept of the archway is at the point $(0, 24)$.

Q: What are the x-intercepts of the archway? A: The x-intercepts of the archway are at the points $(-3, 0)$ and $(8, 0)$.

Q: What is the width of the archway at its base? A: The width of the archway at its base is $11$ units.

Q: What is the height of the archway? A: The height of the archway is $\frac{121}{4}$ units.

Q: How do I find the y-intercept of the archway? A: To find the y-intercept of the archway, set $x = 0$ in the equation $y = -x^2 + 5x + 24$.

Q: How do I find the x-intercepts of the archway? A: To find the x-intercepts of the archway, set $y = 0$ in the equation $y = -x^2 + 5x + 24$ and solve for $x$.

Q: How do I find the width of the archway at its base? A: To find the width of the archway at its base, find the distance between the x-intercepts.

Q: How do I find the height of the archway? A: To find the height of the archway, find the maximum value of the function $y = -x^2 + 5x + 24$.

Q: What is the vertex of the parabola? A: The x-coordinate of the vertex is given by $x = \frac{-b}{2a} = \frac{-5}{2(-1)} = \frac{5}{2}$.

Q: How do I draw a diagram of the archway? A: To draw a diagram of the archway, include the following features:

• The y-intercept at $(0, 24)$
• The x-intercepts at $(-3, 0)$ and $(8, 0)$
• The width of the archway at its base, which is $11$ units
• The height of the archway, which is $\frac{121}{4}$ units
• The equation $y = -x^2 + 5x + 24$ to represent the shape of the archway.

Q: What is the significance of the equation $y = -x^2 + 5x + 24$? A: The equation $y = -x^2 + 5x + 24$ represents the shape of the archway and its key features.

Q: How can I use this information to solve other problems? A: You can use this information to solve other problems involving quadratic equations and parabolas.

Conclusion

In this Q&A article, we have answered frequently asked questions about drawing a diagram of the archway modeled by the equation $y = -x^2 + 5x + 24$. We have provided step-by-step instructions for finding the y-intercept, x-intercepts, width, and height of the archway, as well as drawing a diagram of the archway.