Question

use your graphing calculator to sketch the graph of the function, and then determine the horizontal interval over which the function is increasing.
y = x² - 96
sketch the graph of the function in the viewing window -20,20×-100,100. choose the correct graph below
○ a. ○ b. ○ c. ○ d.

Explanation:

Step1: Analyze the function type

The function given is \(y = x^{2}-96\), which is a quadratic function in the form \(y=ax^{2}+bx + c\) where \(a = 1\), \(b = 0\) and \(c=- 96\). For a quadratic function \(y = ax^{2}+bx + c\), if \(a>0\), the parabola opens upwards. Since \(a = 1>0\), the graph of the function should be an upward - opening parabola.

Step2: Analyze the vertex of the parabola

The x - coordinate of the vertex of a quadratic function \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). Substituting \(a = 1\) and \(b = 0\) into the formula, we get \(x = 0\). To find the y - coordinate of the vertex, we substitute \(x = 0\) into the function \(y=x^{2}-96\), so \(y=0^{2}-96=-96\). So the vertex of the parabola is at \((0,-96)\).

Step3: Match with the given graphs

• Option A: The parabola opens downwards (since it has a maximum point at the vertex), so it does not match as our function has \(a = 1>0\) and should open upwards.
• Option B: The parabola opens downwards, so it is incorrect.
• Option C: The parabola opens upwards and the vertex is at a negative y - value (consistent with \(y=-96\) when \(x = 0\)), so this graph matches the function \(y=x^{2}-96\).
• Option D: The vertex of the parabola in this option seems to be at a non - negative y - value (close to 0 or positive), but our vertex is at \((0,-96)\), so it does not match.

Answer:

C. The graph with the upward - opening parabola (the third graph option)