To solve the problem of selecting the graph that shows the relation x=y2−5 x = y^2 - 5 x=y2−5, we need to understand the shape and orientation of the graph. This equation represents a parabola that opens to the right because x x x is expressed in terms of y y y. The vertex of the parabola is at (−5,0)( -5, 0 )(−5,0).
The given equation is: x=y2−5 x = y^2 - 5 x=y2−5
This is a quadratic equation in terms of y y y. To understand the graph of this equation, we need to analyze its properties.
The equation x=y2−5 x = y^2 - 5 x=y2−5 represents a parabola. Since y y y is squared, the parabola opens horizontally. Specifically, it opens to the right because the coefficient of y2 y^2 y2 is positive.
The vertex of the parabola can be found by setting y=0 y = 0 y=0: x=02−5=−5 x = 0^2 - 5 = -5 x=02−5=−5 So, the vertex is at the point (−5,0)(-5, 0)(−5,0).
To better understand the graph, we can plot a few key points by substituting values for y y y:
When y=1 y = 1 y=1: x=12−5=−4 x = 1^2 - 5 = -4 x=12−5=−4 Point: (−4,1)(-4, 1)(−4,1)
When y=−1 y = -1 y=−1: x=(−1)2−5=−4 x = (-1)^2 - 5 = -4 x=(−1)2−5=−4 Point: (−4,−1)(-4, -1)(−4,−1)
When y=2 y = 2 y=2: x=22−5=−1 x = 2^2 - 5 = -1 x=22−5=−1 Point: (−1,2)(-1, 2)(−1,2)
When y=−2 y = -2 y=−2: x=(−2)2−5=−1 x = (-2)^2 - 5 = -1 x=(−2)2−5=−1 Point: (−1,−2)(-1, -2)(−1,−2)
Using the vertex and the key points, we can sketch the graph. The parabola opens to the right with the vertex at (−5,0)(-5, 0)(−5,0) and passes through the points (−4,1)(-4, 1)(−4,1), (−4,−1)(-4, -1)(−4,−1), (−1,2)(-1, 2)(−1,2), and (−1,−2)(-1, -2)(−1,−2).
The graph that shows the relation x=y2−5 x = y^2 - 5 x=y2−5 is a horizontally opening parabola with the vertex at (−5,0)(-5, 0)(−5,0). The correct graph will reflect this shape and these key points.
The graph is a horizontally opening parabola with vertex at (−5,0). \boxed{\text{The graph is a horizontally opening parabola with vertex at } (-5, 0).} The graph is a horizontally opening parabola with vertex at (−5,0).
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