PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
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Chapter P.5, Problem 45E
Solution

a.

The two equations that are graphed in the given image.

The parabolic graph is y=x2−1 and the graph that begins from horizontal axis is y=3x+4 .

Given:

The graph to solve the equation 3x+4=x2−1 ,

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter P.5, Problem 45E , additional homework tip 1

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, 3x+4=x2−1 , the graphs of each side function is drawn and then located their intersection points as a solution of the 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 y=x2−1 and the graph that begins from horizontal axis is y=3x+4 .

b.

The two equations that are graphed in the given image.

The equation graphed is y=3x+4−x2+1

Given:

The graph to solve the equation 3x+4=x2−1 ,

  PRECALCULUS:...COMMON CORE ED.-W/ACCESS, Chapter P.5, Problem 45E , additional homework tip 2

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, 3x+4=x2−1 , the graphs of the combined equation y=3x+4−x2+1 is drawn and then located the x -intercepts.

Thus, the equation graphed is y=3x+4−x2+1 .

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 3x+4=x2−1 ,

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).

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Chapter P.5, Problem 44E
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Chapter P.5, Problem 46E

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

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