graph the given functions.
41. (y = x^2 + 1)
42. (y = x^3 + 5)
43. (y = x + 2)
44. (v = 9pi h)
45. (a = pi r^2)

Question

Accepted Answer

# Explanation:

⚡ Using what you learned: representing functions (tables, graphs, equations, mappings)

## Step 1: Graph of $ y = x^2 + 1 $

This is a quadratic function. The parent function $ y = x^2 $ is a parabola opening upwards with its vertex at $ (0,0) $. Adding $ 1 $ shifts the entire graph vertically upward by $ 1 $ unit.

*   **Vertex:** $ (0, 1) $
*   **Key Points:** $ (-2, 5) $, $ (-1, 2) $, $ (0, 1) $, $ (1, 2) $, $ (2, 5) $

$$
\begin{array}{c|c}
x & y \
\hline
-2 & 5 \
-1 & 2 \
0 & 1 \
1 & 2 \
2 & 5 \
\end{array}
$$

---

## Step 2: Graph of $ y = x^3 + 5 $

This is a cubic function. The parent function $ y = x^3 $ has an S-shaped curve passing through the origin. Adding $ 5 $ shifts the inflection point vertically upward by $ 5 $ units.

*   **Inflection Point:** $ (0, 5) $
*   **Key Points:** $ (-2, -3) $, $ (-1, 4) $, $ (0, 5) $, $ (1, 6) $, $ (2, 13) $

$$
\begin{array}{c|c}
x & y \
\hline
-2 & -3 \
-1 & 4 \
0 & 5 \
1 & 6 \
2 & 13 \
\end{array}
$$

---

## Step 3: Graph of $ y = x + 2 $

This is a linear function in slope-intercept form $ y = mx + b $, where the slope $ m = 1 $ and the $ y $-intercept is $ (0, 2) $.

*   **$ y $-intercept:** $ (0, 2) $
*   **$ x $-intercept:** $ (-2, 0) $
*   **Key Points:** $ (-2, 0) $, $ (0, 2) $, $ (2, 4) $

$$
\begin{array}{c|c}
x & y \
\hline
-2 & 0 \
0 & 2 \
2 & 4 \
\end{array}
$$

---

## Step 4: Graph of $ V = 9\pi h $

This is a linear function representing volume $ V $ in terms of height $ h $. Since physical dimensions (volume and height) must be non-negative, we restrict the domain to $ h \ge 0 $. The graph is a ray starting at the origin with a steep positive slope of $ 9\pi \approx 28.27 $.

*   **Starting Point:** $ (0, 0) $
*   **Key Points:** $ (0, 0) $, $ (1, 9\pi) $, $ (2, 18\pi) $

$$
\begin{array}{c|c}
h & V \
\hline
0 & 0 \
1 & 9\pi \approx 28.27 \
2 & 18\pi \approx 56.55 \
\end{array}
$$

---

## Step 5: Graph of $ A = \pi r^2 $

This is a quadratic function representing the area $ A $ of a circle in terms of its radius $ r $. Since the radius and area must be non-negative, the domain is restricted to $ r \ge 0 $. The graph is the right half of a parabola starting at the origin.

*   **Starting Point:** $ (0, 0) $
*   **Key Points:** $ (0, 0) $, $ (1, \pi) $, $ (2, 4\pi) $, $ (3, 9\pi) $

$$
\begin{array}{c|c}
r & A \
\hline
0 & 0 \
1 & \pi \approx 3.14 \
2 & 4\pi \approx 12.57 \
3 & 9\pi \approx 28.27 \
\end{array}
$$

# Answer:

Below are the descriptions and key plotting coordinates to construct each graph:

1.  **$ y = x^2 + 1 $**: A parabola opening upwards with its vertex at $ (0, 1) $. Plot points $ (-2, 5) $, $ (-1, 2) $, $ (0, 1) $, $ (1, 2) $, and $ (2, 5) $, then connect them with a smooth curve.
2.  **$ y = x^3 + 5 $**: An S-shaped cubic curve with an inflection point at $ (0, 5) $. Plot points $ (-2, -3) $, $ (-1, 4) $, $ (0, 5) $, $ (1, 6) $, and $ (2, 13) $, then connect them with a smooth curve.
3.  **$ y = x + 2 $**: A straight line passing through the $ y $-intercept $ (0, 2) $ and $ x $-intercept $ (-2, 0) $. Use a straightedge to draw a line through these points.
4.  **$ V = 9\pi h $**: A straight ray starting at the origin $ (0, 0) $ and extending into the first quadrant (since $ h \ge 0 $) with a slope of $ 9\pi \approx 28.27 $.
5.  **$ A = \pi r^2 $**: The right half of a parabola starting at the origin $ (0, 0) $ and curving upwards into the first quadrant (since $ r \ge 0 $). Plot points $ (0, 0) $, $ (1, \pi \approx 3.14) $, and $ (2, 4\pi \approx 12.57) $.

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QUESTION IMAGE

Question

Explanation:

Step 1: Graph of \( y = x^2 + 1 \)

Step 2: Graph of \( y = x^3 + 5 \)

Step 3: Graph of \( y = x + 2 \)

Step 4: Graph of \( V = 9\pi h \)

Step 5: Graph of \( A = \pi r^2 \)

Answer: