Martha wrote an example of a quadratic function for a homework assignment. The function she wrote is shown:
[tex]f(x)=5 x^3+2 x^2+7 x-3[/tex]
What possible changes can Martha make to correct her homework assignment? Choose two correct answers.
A. The exponent on the first term, [tex]5 x^3[/tex], can be changed to a 2 and then combined with the second term, [tex]2 x^2[/tex].
B. The exponent on the second term, [tex]2 x^2[/tex], can be changed to a 3 and then combined with the first term, [tex]5 x^3[/tex].
C. The constant, -3, can be changed to a variable.
D. The first term, [tex]5 x^3[/tex], can be eliminated.
E. The [tex]7 x[/tex] can be eliminated.
Answer (2)
Changing the exponent on the first term to 2 and combining it with the second term results in a quadratic function.
Eliminating the first term results in a quadratic function.
The two possible changes are: changing the exponent on the first term to 2 and eliminating the first term.
The possible changes are
Explanation
Understanding Quadratic Functions We are given the function f ( x ) = 5 x 3 + 2 x 2 + 7 x − 3 and we need to identify two changes that would make it a quadratic function. A quadratic function has the general form a x 2 + b x + c , where a , b , and c are constants and a e q 0 . This means the highest power of x must be 2.
Evaluating Option 1 Option 1: Change the exponent on the first term, 5 x 3 , to a 2 and combine it with the second term, 2 x 2 .
If we change the exponent of the first term to 2, we get 5 x 2 . Combining this with the second term, 2 x 2 , gives us 5 x 2 + 2 x 2 = 7 x 2 . The resulting function would be f ( x ) = 7 x 2 + 7 x − 3 , which is a quadratic function. So, this is a valid change.
Evaluating Option 2 Option 2: Change the exponent on the second term, 2 x 2 , to a 3 and combine it with the first term, 5 x 3 .
If we change the exponent of the second term to 3, we get 2 x 3 . Combining this with the first term, 5 x 3 , gives us 5 x 3 + 2 x 3 = 7 x 3 . The resulting function would be f ( x ) = 7 x 3 + 7 x − 3 , which is a cubic function, not a quadratic function. So, this is not a valid change.
Evaluating Option 3 Option 3: Change the constant, -3, to a variable. Changing the constant to a variable, say x , would result in f ( x ) = 5 x 3 + 2 x 2 + 7 x − x = 5 x 3 + 2 x 2 + 6 x , which is a cubic function, not a quadratic function. So, this is not a valid change.
Evaluating Option 4 Option 4: Eliminate the first term, 5 x 3 .
If we eliminate the first term, the function becomes f ( x ) = 2 x 2 + 7 x − 3 , which is a quadratic function. So, this is a valid change.
Evaluating Option 5 Option 5: Eliminate the 7 x term. If we eliminate the 7 x term, the function becomes f ( x ) = 5 x 3 + 2 x 2 − 3 , which is a cubic function, not a quadratic function. So, this is not a valid change.
Conclusion Therefore, the two possible changes Martha can make to correct her homework assignment are:
Change the exponent on the first term, 5 x 3 , to a 2 and then combined with the second term, 2 x 2 .
The first term, 5 x 3 , can be eliminated.
Examples
Quadratic functions are used in many real-world applications, such as modeling the trajectory of a ball, designing parabolic mirrors, and optimizing business processes. Understanding how to identify and manipulate quadratic functions is a fundamental skill in algebra and calculus. For example, if you were designing a bridge, you might use a quadratic function to model the curve of the arch. Or, if you were trying to maximize the profit from selling a product, you might use a quadratic function to model the relationship between price and demand.
Martha can correct her function by either changing the exponent on the first term, 5 x 3 , to 2 and combining it with the second term, or by eliminating the first term completely. Both changes result in a quadratic function. Therefore, the correct options are A and D.
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