Concept explainers
a.
The two equations that are graphed in the given image.
The parabolic graph is
Given:
The graph to solve the equation
Concept Used:
- The graph of a quadratic function is U shaped.
- The graph of an even order radical function can take values that are greater than or equal to 0 only.
Calculation:
In the given graph, note that there are two graphs, one is an upward opening parabola and the other graph is starting from the horizontal axis and moving upward.
To solve the given equation,
Since the left hand side expression is an square root function, its graph will start from horizontal axis and move upward and the right side expression is a quadratic function, so its graph will be a parabola.
Thus, the parabolic graph is
b.
The two equations that are graphed in the given image.
The equation graphed is
Given:
The graph to solve the equation
Concept Used:
- The graph of a quadratic function is U shaped.
- The graph of an even order radical function can take values that are greater than or equal to 0 only.
Calculation:
In the given graph, note that there is only one graph and it intersecting the horizontal axis at two points.
To solve the given equation,
Thus, the equation graphed is
c.
How the intersection point in figure (a) related to the x -intercept in figure (b).
The intersection points in figure (a) are same as the x -intercepts in figure (b).
Given:
The graphs to solve the equation
Concept Used:
- The graph of a quadratic function is U shaped.
- The graph of an even order radical function can take values that are greater than or equal to 0 only.
Calculation:
From the figure (a) and figure (b), note that the intersection points of figure (a) are exactly the same as the x -intercepts of the figure (b) as they both giving the solutions of the same equation.
Thus, the intersection points in figure (a) are same as the x -intercepts in figure (b).
Chapter P Solutions
EBK PRECALCULUS:GRAPHICAL,...-NASTA ED.
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