Function
f
is defined by
f
(
x
)
=
(
x
−
5
)
2
+
1
.
Function
g
is defined by
g
(
x
)
=
f
(
x
)
−
2
.
Which quadrants in the
x
y
–plane contain points on the graph of
y
=
g
(
x
)
?
Wrong again.
g(x) = (x-5)^2 - 1 has its vertex at (5,-1)
It has x-intercepts at x=4 and x=6
So there are no points in QIII for this graph.
You are correct, I apologize for the mistake. The correct description is as follows:
The graph of y = g(x) does not contain any points in quadrant III of the x-y plane. This is because the graph of g(x) is a parabola that opens upward and its vertex is located at (5, -1). The x-intercepts of g(x) are at x = 4 and x = 6.
Since the x values of the points in quadrant III are negative, and g(x) only has x-intercepts at positive x values, there are no points on the graph of y = g(x) in quadrant III.
The graph of y = g(x) contains points in all quadrants of the x-y plane.
To determine this, we can consider the transformation of the graph of f(x) into g(x). The graph of f(x) is a parabola that opens upward and is shifted 5 units to the right and 1 unit up from the vertex of the standard parabola y = x^2.
When we subtract 2 from f(x) to obtain g(x), the entire graph of f(x) is shifted downward by 2 units. So, the graph of g(x) will still be present in all quadrants of the x-y plane, just shifted downward by 2 units from the graph of f(x).