Problem 61 - Algebra and Trigonometry Real Mathematics, Real People

Question

Sketch the graph of the piecewise-defined function by hand. $$f(x)=\left\\{\begin{array}{ll} 2 x+1, & x \leq-1 \\ x^{2}-2, & x>-1 \end{array}\right.$$

Step-by-Step Solution

Answer

The function consists two pieces which are a line with gradient 2 going to the left from \((-1, -1)\) and a parabola shifted downwards by 2 units starting from just right to point \((-1, -1)\).

Step 1: Understand the function

Firstly, determine the equations for each 'piece' of the function. Our piecewise function consists of two functions, \(2x+1\) when \(x \leq -1\) and \(x^2 - 2\) when \(x > -1\). Therefore we have two parts to sketch - a linear function with gradient 2 and y intercept 1, and a quadratic function that is translated down by 2.

Step 2: Sketch the first piece

The first function applies for \(x \leq -1\). So, when \(x = -1\) get the point \((-1, y)\) on the graph. Substitute \(x = -1\) in the equation to get \(y = 2*-1 + 1 = -1\). So, when \(x = -1, y = -1\). Start from point \((-1, -1)\) and sketch the line with a gradient 2 to the left.

Step 3: Sketch the second piece

The second function applies for \(x > -1\). So, start from \(x = -1\), but in this part of function, the 'x' starts from just greater than -1. This time, substitute \(x = - 1\) in the equation \(x^2 - 2\) to get \(y = -1^2 - 2 = -1\). So, begin from just right to the point \((-1, -1)\) and sketch the curve \(x^2 - 2\) upwards for \(x > -1\).

Key Concepts

Sketching Graphs

Graph sketching is an essential skill in understanding and visualizing mathematical concepts, especially for piecewise-defined functions. For the given exercise, to sketch the graph of the function
\[ f(x) = \begin{cases} 2x+1, & x \leq -1 \ x^2 - 2, & x > -1 \end{cases} \]
you must comprehend the behavior and attributes of each 'piece.' Begin by identifying the domain for each equation and the corresponding range. With that information, here are steps you could follow:

• Identify pivot points: Find the key coordinates where the function changes its definition. In our case, that is where \(x = -1\). Calculate the function output at this point for both equations.
• Draw each function segment: For the linear part \(2x+1\), plot the line which should continue indefinitely to the left of \(x = -1\) as there is no lower limit to its domain. For the quadratic part \(x^2 - 2\), start just the right of \(x = -1\) and draw the curve accordingly for its domain \(x > -1\).
• Ensure continuity or discontinuity is accurate: Depending on the function, a seamless transition or a jump may occur at the pivot point. Verify if the pieces connect or not, keeping in mind the domain limitations.

Graph paper can be of great assistance here. Always label your axes, mark the pivot point clearly, and use a ruler for the linear part and smooth, consistent curves for the quadratic part.

Linear Function

A linear function is of the form \(y = mx + b\), where \(m\) represents the slope or gradient of the line, and \(b\) indicates the y-intercept, which is the point where the line crosses the y-axis. In our function
\[ f(x) = 2x + 1 \text{ for } x \leq -1 \]
the slope (\(m\)) is 2, which means the line rises by 2 units vertically for each unit it moves horizontally to the right. Conversely, it falls by 2 units if you move to the left. The y-intercept (\(b\)) is 1, indicating that our line intersects the y-axis at the point \((0, 1)\). However, because our equation only applies when \(x \leq -1\), the line would start at \((-1, -1)\) and continue left, not touching the y-axis. To ensure your sketch is correct, check that it adheres to the gradient by making use of the rise over run method or by plotting additional points and connecting them.

Quadratic Function

Quadratic functions have the form \(y = ax^2 + bx + c\), describing a parabola shaped curve. The coefficient \(a\) determines the direction and width of the parabola, while \(b\) and \(c\) influence its position on the coordinate plane.
For the second part of our piecewise-defined function, we have
\[ f(x) = x^2 - 2 \text{ for } x > -1 \]
The coefficient \(a\) is positive (\(a = 1\)), which signals that the parabola opens upwards. The lack of a \(bx\) term means it is symmetrical about the y-axis, and \(c = -2\) translates the parabola two units down. The vertex of this parabola is the lowest point \((0, -2)\), but because it's part of the segment where \(x > -1\), for sketching, start at the point just to the right of \((-1, -1)\) and continue the curve upwards. Remember, the wider the parabola, the smaller the coefficient \(a\) is. In our case, since \(a\) is 1, the parabola will appear quite standard in width, becoming steeper as \(x\) increases.