It looks like you've uploaded an image containing a math problem related to
conic sections, specifically involving
parabolas. The problem is about a
parabolic arch with a given width and height, and there are several questions about its equation, dimensions, and properties.
Since I can't view images directly, I’ll solve the problem based on the text you provided. Let's reconstruct and solve it step by step.
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📌 Problem Summary:
A
parabolic arch has:
- Width: 10 meters (from one end to the other)
- Height: 5 meters at the center
The parabola opens
downward, and we're to find:
1. The equation of the parabola
2. The height at a specific point (e.g., 3 meters from the center)
3. The width at a certain height
4. The maximum height (already given)
We'll assume the vertex of the parabola is at the top center, so the vertex is at $(0, 5)$, and the arch spans from $x = -5$ to $x = 5$, touching the ground at both ends ($y = 0$).
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✔ Step 1: Find the Equation of the Parabola
General form of a downward-opening parabola with vertex at $(h, k)$:
$$
y = a(x - h)^2 + k
$$
Vertex is at $(0, 5)$, so:
$$
y = a x^2 + 5
$$
Now use the fact that at $x = 5$, $y = 0$ (since the arch touches the ground at the edge):
$$
0 = a(5)^2 + 5 \Rightarrow 0 = 25a + 5 \Rightarrow 25a = -5 \Rightarrow a = -\frac{1}{5}
$$
✔ So the equation is:
$$
\boxed{y = -\frac{1}{5}x^2 + 5}
$$
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✔ Question 1: What is the equation of the parabola?
>
Answer: $ y = -\frac{1}{5}x^2 + 5 $
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✔ Question 2: How high is the arch 3 meters from the center?
Plug $x = 3$ into the equation:
$$
y = -\frac{1}{5}(3)^2 + 5 = -\frac{9}{5} + 5 = -1.8 + 5 = 3.2
$$
✔ So the height is
3.2 meters.
>
Answer: $ \boxed{3.2} $ meters
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✔ Question 3: What is the width of the arch at a height of 2 meters?
Set $y = 2$ and solve for $x$:
$$
2 = -\frac{1}{5}x^2 + 5 \Rightarrow -\frac{1}{5}x^2 = -3 \Rightarrow x^2 = 15 \Rightarrow x = \pm\sqrt{15}
$$
So the width is the distance between $-\sqrt{15}$ and $\sqrt{15}$:
$$
\text{Width} = 2\sqrt{15} \approx 2 \times 3.873 = 7.746 \text{ meters}
$$
✔ Answer: $ \boxed{2\sqrt{15}} $ meters or approximately $ \boxed{7.75} $ meters
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✔ Question 4: What is the maximum height of the arch?
Given: The vertex is at $y = 5$
✔ Answer: $ \boxed{5} $ meters
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✔ Final Answers:
1. Equation: $ \boxed{y = -\frac{1}{5}x^2 + 5} $
2. Height at 3 m from center: $ \boxed{3.2} $ meters
3. Width at height 2 m: $ \boxed{2\sqrt{15}} $ meters ≈ $ \boxed{7.75} $ meters
4. Maximum height: $ \boxed{5} $ meters
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Let me know if your image had different values (e.g., different width/height), and I can adjust accordingly!
Parent Tip: Review the logic above to help your child master the concept of reading comprehension worksheet 10th grade.
