Concept explainers
a.
To predict: The changes in the graph of the function y=a(x−h)2+k , where a=1 , h=3 and k=−2 , when a changes to −3 .
The revised function y=−3(x−3)2−2 opens in oppositive direction as that of the original function and is compressed in y-direction by a factor of 3.
Given information:
The function y=a(x−h)2+k , where a=1 , h=3 and k=−2 is
The value a changes to −3 .
Concept used:
For a function of the type y=C⋅f(x) , if C>1 , it compresses the function in the x-direction and if 0<C<1 , it stretches the function in x-direction.
Calculation:
The function y=(x−3)2−2 is a quadratic function with a=1 . This represents an upward parabola.
Multiply the term (x−3)2 with negative. Then, the function becomes, y=−(x−3)2−2 . This shows that the revised function is a downward parabola.
Now, multiply the term −(x−3)2 by 3. So, the function becomes −3(x−3)2 .
Since 3>1 , the function y=−3(x−3)2−2 is compressed in y-direction.
Therefore, the function y=−(x−3)2−2 is downward parabola and is compressed in y-direction to get the graph of the function y=−3(x−3)2−2 .
Now, graph the original function and the revised function using a graphing calculator.
The graph shows that the prediction about the revised function is correct.
Conclusion:
The revised function y=−3(x−3)2−2 opens in oppositive direction as that of the original function and is compressed in y-direction by a factor of 3.
b.
To predict: The changes in the graph of the function y=a(x−h)2+k , where a=1 , h=3 and k=−2 , when h changes to −1 .
The revised function is shifted 4 units to the left of that of the original function.
Given information:
The function y=a(x−h)2+k , where a=1 , h=3 and k=−2 is
The value h changes to −1 .
Concept used:
For a function of the type y=f(x+C) , if C>0 , the original function f(x) shifts C units to the left and if C<0 , the original function f(x) shifts C units to the right.
Calculation:
The function y=(x−3)2−2 has vertex at the point (3,−2) .
When h changes to −1 , the function becomes y=(x+1)2−2 . This has the vertex at the point (−1,−2) .
That means, the revised function is shifted 4 units to the left of that of the original function.
Now, graph the original function and the revised function using a graphing calculator.
The graph shows that the revised function is shifted 4 units left to that of the original function.
Conclusion:
The revised function is shifted 4 units to the left of that of the original function.
c.
To predict: The changes in the graph of the function y=a(x−h)2+k , where a=1 , h=3 and k=−2 , when k changes to 2 .
The revised function is shifted 4 units to the left of that of the original function.
Given information:
The function y=a(x−h)2+k , where a=1 , h=3 and k=−2 is
The value k changes to 2 .
Concept used:
For a function of the type y=f(x+C) , if C>0 , the original function f(x) shifts C units upward and if C<0 , the original function f(x) shifts C units downward.
Calculation:
The function y=(x−3)2−2 has vertex at the point (3,−2) .
When k changes to 2 , the function becomes y=(x−3)2+2 . This has the vertex at the point (3,2) .
That means, the revised function is shifted 4 units upwards to that of the original function.
Now, graph the original function and the revised function using a graphing calculator.
The graph shows that the revised function is shifted 4 units upwards to that of the original function.
Conclusion:
The revised function is shifted 4 units upwards to that of the original function.
Chapter 1 Solutions
Algebra 2: New York Edition (holt Mcdougal Larson Algebra 2)
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