Question

graph the function.
$p(x) = x^2 - 1$

Explanation:

Step1: Identify the function type

The function \( p(x) = x^2 - 1 \) is a quadratic function, which has a parabolic graph. The general form of a quadratic function is \( y = ax^2 + bx + c \), here \( a = 1 \), \( b = 0 \), \( c = - 1 \). Since \( a>0 \), the parabola opens upwards.

Step2: Find the vertex

For a quadratic function \( y = ax^2+bx + c \), the x - coordinate of the vertex is \( x=-\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = 0 \), we get \( x = 0 \). When \( x = 0 \), \( y=p(0)=0^2 - 1=-1 \). So the vertex of the parabola is at \( (0,-1) \).

Step3: Find the x - intercepts

Set \( y = 0 \), then \( x^2-1=0 \). Using the difference of squares formula \( a^2 - b^2=(a + b)(a - b) \), we have \( (x + 1)(x - 1)=0 \). So \( x+1 = 0 \) or \( x - 1=0 \), which gives \( x=-1 \) or \( x = 1 \). The x - intercepts are \( (-1,0) \) and \( (1,0) \).

Step4: Sketch the graph

Plot the vertex \( (0,-1) \), the x - intercepts \( (-1,0) \) and \( (1,0) \). Since the parabola opens upwards, draw a smooth curve passing through these points.

(If we were to use the given graphing tool: First, select the parabola tool (the "U" shaped icon). Then, plot the vertex at (0, - 1) and use the x - intercepts or other points to define the shape of the parabola. The parabola should open upwards with vertex at (0, - 1) and pass through (-1, 0) and (1, 0).)

Answer:

To graph \( p(x)=x^2 - 1 \):

1. Recognize it's an upward - opening parabola (since the coefficient of \( x^2\) is positive) with vertex at \( (0,-1) \), x - intercepts at \( (-1,0) \) and \( (1,0) \).
2. Using the graphing tool: Select the parabola (U - shaped) tool, plot the vertex \( (0,-1) \), and use the x - intercepts or the shape of the parabola to draw the graph. The graph is a parabola opening upwards with vertex at \( (0, - 1) \) and x - intercepts at \( (-1,0) \) and \( (1,0) \).