Question

use your graphing calculator to sketch the graph of the function, and then determine each of the coordinates of the x-intercepts for the function, if they exist.
$y = x^2 - 45$
sketch the graph of the function in the viewing window $-10,10 \times -50,10$. choose the correct graph below.
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.
\\(\bigcirc\\) c.
\\(\bigcirc\\) d.

Explanation:

1. First, analyze the function \(y = x^{2}-45\). This is a quadratic function in the form \(y = ax^{2}+bx + c\), where \(a = 1\), \(b = 0\), and \(c=- 45\). Since \(a=1>0\), the parabola opens upward.
2. Next, find the vertex of the parabola. The x - coordinate of the vertex of a parabola \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\). Substituting \(a = 1\) and \(b = 0\), we get \(x = 0\). Substituting \(x = 0\) into the function, we get \(y=0^{2}-45=-45\). So the vertex is at \((0,-45)\).
3. Now, consider the graphs:

• Graph A and C open downward (since they have a maximum point at the vertex), so they can be eliminated because our function has \(a = 1>0\) and should open upward.
• Graph D: The vertex of the parabola in Graph D seems to be at a non - negative y - value (close to 0 or positive), but our vertex is at \((0,-45)\).
• Graph B: It opens upward, and the vertex is at a negative y - value (consistent with \(y=-45\) when \(x = 0\)), so it is the correct graph.

Answer:

B. The graph with the upward - opening parabola (the second graph)