Problem 30

Question

Sketch the graphs of the function $g(x)=f(x)+C$ for $C=-2, C=0,$ and $C=3$ on the same set of coordinate axes. $$ f(x)=\sqrt{x} $$

Step-by-Step Solution

Verified

Answer

For the function $g(x) = f(x)+c$, it is noted that the function is shifted up or down depending on the value of $c$. For $c=0$, no shift happens and the function remains the same as $f(x)$. For $c=-2$, the function is shifted down by 2 units and for $c=3$, the function is shifted upwards by 3 units.

1Step 1: Sketching function for $C=0$

With $C=0$, $g(x) = f(x) = \sqrt{x}$. Draw the graph of this function, which starts from the origin (0, 0) and increases slowly. It is a curve in the first quadrant.

2Step 2: Sketching function for $C=-2$

With $C=-2$, the function $g(x) = \sqrt{x} - 2$. The graph of this function is the same as that of $f(x)$, but shifted down by 2 units. It starts from the point (0, -2) and increases slowly.

3Step 3: Sketching function for $C=3$

With $C=3$, the function $g(x) = \sqrt{x} + 3$. The graph of this function is the same as $f(x)$ but it is shifted up by 3 units. It starts from the point (0, 3) and increases slowly.

Key Concepts

Function TranslationSqrt FunctionCoordinate Plane

Function Translation

Function translation involves altering the position of a graph on the coordinate plane. This can be done vertically or horizontally by adding or subtracting a constant value to the function.
In the case of the sqrt function represented by $ f(x) = \sqrt{x} $, adding a constant $ C $ results in a vertical shift.

When $ C $ is positive, the function shifts upward.
When $ C $ is negative, the function shifts downward.

For example, if we have $ g(x) = \sqrt{x} + 3 $, it means the graph will move 3 units up. Conversely, $ g(x) = \sqrt{x} - 2 $ translates the graph 2 units down.
Function translation does not affect the shape of the graph, only its position. Understanding this concept helps in predicting how graphs will move on the coordinate plane when constants are adjusted.

Sqrt Function

The square root function, denoted by $ f(x) = \sqrt{x} $, is a popular function due to its interesting properties. It is a specific type of function known as a radical function. Here are some key points about the square root function:

The domain is all non-negative numbers $ x \geq 0 $, since we cannot take the square root of negative numbers in real number mathematics.
The range is $ y \geq 0 $, as the square root of any number is non-negative.
The graph starts at the origin $(0, 0)$ and curves gradually upward, moving towards infinity as $ x $ increases.

This curve signifies that the function grows, but at a decreasing rate, as the input value $ x $ becomes larger. Every increase in $ x $ results in a smaller increase in the corresponding $ y $ value, creating a gentle slope that is characteristic of the sqrt function.
Understanding the behavior of the square root function is crucial when learning about graph transformations and translations as it lays the foundation for more complex shifts and stretches.

Coordinate Plane

The coordinate plane is a two-dimensional plane formed by the intersection of a vertical line (y-axis) and a horizontal line (x-axis). These axes allow us to graph functions and form the basis for understanding graph transformations such as the function translation.

The x-axis (horizontal) allows us to plot the independent variable, often representing the input for functions like $ f(x)=\sqrt{x} $.
The y-axis (vertical) represents the dependent variable or the output of functions.

Every point on the coordinate plane is represented by an ordered pair $(x, y)$, where $ x $ is the horizontal component and $ y $ is the vertical component. When sketching graphs, correctly placing points and understanding shifts in relation to these axes is essential. For example, with the function $ g(x) = \sqrt{x} + C $ where $ C $ is a constant, the entire graph rises or lowers without changing its shape or orientation, just its position on the coordinate plane. Understanding this setup aids in better comprehension and visualization of various graph transformations.

Problem 30

Problem 31

Other exercises in this chapter

Problem 30

Solve the differential equation. Use a graphing utility to graph three solutions, one of which passes through the given point. $$ \frac{d r}{d t}=\frac{\sec ^{2

View solution

Problem 30

In Exercises $27-30,$ find the limit of $s(n)$ as $n \rightarrow \infty$ $$ s(n)=\frac{1}{n^{2}}\left[\frac{n(n+1)}{2}\right] $$

View solution

Problem 31

In Exercises 31 and $32,$ show that the function satisfies the differential equation. $y=a \sinh x$ $y^{\prime \prime \prime}-y^{\prime}=0$

View solution

Problem 31

Use the specified substitution to find or evaluate the integral. $$ \begin{array}{l} \int \sqrt{e^{t}-3} d t \\ u=\sqrt{e^{t}-3} \end{array} $$

View solution