Problem 40
Question
33–48 ? Sketch the graph of the function, not by plotting points, but by starting with the graph of a standard function and applying transformations. $$ y=2-\sqrt{x+1} $$
Step-by-Step Solution
Verified Answer
The graph is a horizontally shifted left half of an inverted square root, shifted up by 2 units.
1Step 1: Identify the Base Function
The function we begin with is the square root function, which is given by \( y = \sqrt{x} \). This is our base function and will be the starting point for applying various transformations.
2Step 2: Apply the Horizontal Shift
The given function is \( y = 2 - \sqrt{x+1} \). The term \( x+1 \) indicates a horizontal shift. To understand how this affects the graph, realize that it shifts the graph of \( \sqrt{x} \) to the left by 1 unit. This change directs us to adjust the way \( x \) is interpreted in the modified function.
3Step 3: Apply the Vertical Reflection and Scaling
The presence of the negative sign in front of \( \sqrt{x+1} \) in \( y = 2 - \sqrt{x+1} \) implies reflecting the graph of the shifted function \( \sqrt{x+1} \) across the x-axis. Thus, we transform \( \sqrt{x+1} \) to \( -\sqrt{x+1} \).
4Step 4: Apply the Vertical Shift
Finally, the term \( 2 - \sqrt{x+1} \) causes a vertical shift up by 2 units. This is evident from adding 2 to the function \( -\sqrt{x+1} \). Every point on the graph is moved 2 units upwards, completing the transformation.
5Step 5: Sketch the Graph
Combine all the transformations step-by-step on the graph of \( \sqrt{x} \): shift left by 1 unit, reflect across the x-axis, then shift up by 2 units. The resulting curve passes through the point (0, 2), descends leftwards, and resembles an inverted half-parabola.
Key Concepts
Understanding the Square Root FunctionApplying a Horizontal ShiftUnderstanding Vertical ReflectionExecuting a Vertical Shift
Understanding the Square Root Function
The square root function, denoted by \( y = \sqrt{x} \), is a fundamental mathematical function. Its graph is characterized by a gentle, upward slope starting from the origin, (0,0). The curve only exists in the first quadrant of the Cartesian plane because you can’t take the square root of a negative number in the realm of real numbers. This graph will serve as our base or "parent" function for transformations.
- The domain of \( y = \sqrt{x} \) is \( x \geq 0 \).
- The range is \( y \geq 0 \).
Applying a Horizontal Shift
Horizontal shifts involve moving the graph left or right on the Cartesian plane. In the equation \( y = 2 - \sqrt{x+1} \), the term \( x+1 \) indicates a horizontal shift. We interpret this as the graph of \( \sqrt{x} \) being shifted to the left by 1 unit. Intuitively, you can think of it as preparing for \( x \), thus modifying the input before the square root operation.
- A positive inside the function, like \( x+1 \), shifts the graph left.
- A negative value, such as \( x-1 \), would shift it right.
Understanding Vertical Reflection
A vertical reflection changes the direction in which the graph opens. For the square root function, \( y = -\sqrt{x+1} \), the presence of a negative sign outside implies a reflection across the x-axis. This means the range of the function will change in direction: instead of going upwards, it will flip, moving downwards.
- The graph, originally increasing upwards, will now appear as it descends from its starting point.
- This does not affect the domain.
Executing a Vertical Shift
A vertical shift involves moving the graph up or down on the plane. In the function \( y = 2 - \sqrt{x+1} \), the +2 signals a vertical shift upwards. This means that every point on the graph of \( -\sqrt{x+1} \) is lifted by 2 units.
- If the number outside is positive, the shift is upwards.
- If it's negative, the graph moves down.
Other exercises in this chapter
Problem 40
Find a function whose graph is a parabola with vertex \((3,4)\) and that passes through the point \((1,-8) .\)
View solution Problem 40
Sketch the graph of the piecewise defined function. $$ f(x)=\left\\{\begin{array}{ll}{1-x} & {\text { if } x
View solution Problem 40
Find the inverse function of \(f\). \(f(x)=5-4 x^{3}\)
View solution Problem 40
Find the domain of the function. $$ f(x)=x^{2}+1, \quad 0 \leq x \leq 5 $$
View solution