Correct Answer
Quadrants I, II, and IV should be selected.
Plot of g of x \( g(x) \)
Explanation
Let's analyze the given functions and determine which quadrants the graph of y equals g of x \( y = g(x) \) passes through. The original functions are:f of x equals open parenthesis x minus 5 close parenthesis squared plus 1, andg of x equals f of x minus 2.
$$ \begin{align}f(x) & = (x - 5)^2 + 1 \\g(x) & = f(x) - 2 \end{align} $$
Substituting f of x \( f(x) \) to get the equation for the graph of y equals g of x \( y = g(x), \) we getg of x equals open parenthesis x minus 5 close parenthesis squared plus 1 minus 2, which equals open parenthesis x minus 5 close parenthesis squared minus 1.
$$ g(x) = (x - 5)^2 + 1 - 2 = (x - 5)^2 - 1 $$
So, the equation for the graph is:
y equals g of x, which equals open parenthesis x minus 5 close parenthesis squared minus 1. $$ y = g(x) = (x - 5)^2 - 1 $$
This function is a parabola that opens upwards because the coefficient of the open parenthesis x minus 5 close parenthesis squared \( (x - 5)^2 \) term is positive. The vertex form of the parabola is y equals open parenthesis x minus h close parenthesis squared plus k, \( y = (x - h)^2 + k, \) where coordinates h, k \( (h, k) \) is the vertex. Here, h equals 5 \( h = 5 \) and k equals negative 1, \( k = -1, \) so the vertex of the parabola is at coordinates 5 negative 1, \( (5, -1), \) which is in Quadrant IV.
Next, you need to determine where the parabola lies. Since it open upward and has a vertex below the x-axis, one way to do that is to determine the x-intercepts, which can be accomplished by setting the value of the function to 0 and solving for x:0 equals open parenthesis x minus 5 close parenthesis squared minus 1.
$$ 0 = (x - 5)^2 - 1 $$
open parenthesis x minus 5 close parenthesis squared equals 1 $$ (x - 5)^2 = 1 $$
x minus 5 equals plus or minus 1 $$ x - 5 = \pm 1 $$
So, x equals 6 \( x = 6 \) and x equals 4, \( x = 4, \) and the x-intercepts are coordinates 6, 0 \( (6, 0) \) and coordinates \( (4, 0). \)
So far, we know there are points in Quadrant IV, including the vertex of the parabola, and in Quadrants I and II, since the parabola opens up. Do we have any points in Quadrant III? To determine this, we need to know whether the y-intercept is above or below the x-axis. The y-intercept can be found by setting x equals 0 \( x = 0 \) and evaluating the function:y equals open parenthesis 0 minus 5 close parenthesis squared minus 1, which equals 25 minus 1, which equals 24.
$$ y = (0 - 5)^2 - 1 = 25 - 1 = 24 $$
Because the vertex is at
coordinates 5, negative 1, \( (5, -1), \) by the time the parabola crosses the y-axis at 24, it is already above the x-axis. So there are no points in Quadrant III. We also know from this that, since the parabola crosses the y-axis at 24, it crosses from Quadrant I to Quadrant II and continues in Quadrant II from there up to infinity. That's why we need to select Quadrant II as a correct answer.
