Problem 27 - College Algebra

Question

For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. $$ f(t)=(t+1)^{2}-3 $$

Step-by-Step Solution

Answer

The function is a parabola shifted left by 1 and down by 3 from \( t^2 \).

Step 1: Identify the Toolkit Function

The given function is \( f(t) = (t+1)^2 - 3 \). The toolkit function it is derived from is the basic quadratic function \( g(t) = t^2 \). Recognize this as a familiar parabola that opens upwards with its vertex at the origin (0,0).

Step 2: Determine Horizontal Shifts

The function \( f(t) = (t+1)^2 - 3 \) includes a horizontal transformation. The term \((t+1)\) indicates a shift to the left by 1 unit. This moves the vertex of the parabola from the origin to \((-1, 0)\).

Step 3: Determine Vertical Shifts

The constant \(-3\) in the function indicates a vertical shift downward by 3 units. The new position of the vertex becomes \((-1, -3)\).

Step 4: Sketch the Graph

Start by plotting the new vertex at \((-1, -3)\). Since the function is derived from the quadratic \( t^2 \), the parabola maintains its shape and opens upwards. Draw the parabola ensuring it is symmetric around the line \(t = -1\), with the vertex being the lowest point of the graph.

Key Concepts

Graph of Quadratic Function

To understand function transformations, it's essential to start with the graph of a quadratic function. A quadratic function, typically represented as \( f(x) = ax^2 + bx + c \), graphs into a shape known as a parabola. This parabola can either open upwards or downwards depending on the sign of \( a \):

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

For the function \( f(t) = (t+1)^2 - 3 \), the graph is a parabola that opens upwards because the leading term \( (t+1)^2 \) has an implicit positive coefficient (\( a = 1 \)). The graph will be symmetric about its vertex, making it important to clearly determine transformations to accurately sketch its position.

Toolkit Functions

Toolkit functions are basic functions from which more complex functions can be derived through transformations. They serve as a starting point for identifying changes made to a function. Some common toolkit functions include linear, quadratic, cubic, exponential, and trigonometric functions. The quadratic toolkit function \( g(x) = x^2 \) is especially important for this exercise.For our specific function \( f(t) = (t+1)^2 - 3 \), the toolkit function is \( g(t) = t^2 \), which is simply a basic quadratic equation with its vertex at the origin (0,0). By altering this basic form through transformations such as translations, we modify where and how the graph displays on a coordinate plane.

Horizontal Shifts

A horizontal shift in a graph is a transformation that moves the graph to the left or right along the x-axis. This occurs by altering the variable inside the function's input, commonly seen as \( f(x) = (x - h)^2 \). The value of \( h \) determines the direction:

• A negative \( h \) shifts the graph to the right by \( |h| \) units.
• A positive \( h \) shifts the graph to the left by \( |h| \) units.

For the function \( f(t) = (t+1)^2 - 3 \), the term \((t+1)\) indicates a shift to the left by 1 unit because you can interpret it as \( t - (-1) \). This moves the vertex of the parabola from its original position at (0,0) to (-1,0).

Vertical Shifts

Vertical shifts move a graph up or down along the y-axis, which is accomplished by adding or subtracting a constant from the function. This doesn't affect the shape of the graph but changes its position vertically.

• Adding a positive constant shifts the graph upwards.
• Adding a negative constant shifts the graph downwards.

In the function \( f(t) = (t+1)^2 - 3 \), the \(-3\) indicates a vertical shift downwards by 3 units. So, the vertex moves from (-1,0) to (-1,-3). This means the entire parabola appears lower on the graph, but its overall shape and symmetry remain unchanged.