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The table below provides values of quadratic function f for some value
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23 Feb 2022, 05:45
23 Feb 2022, 05:45
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The table below provides values of quadratic function f for some values x. Which of the following equations could define f(x)?
GRe The table below.jpg [ 12.83 KiB | Viewed 1768 times ]
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
Attachment:
GRe The table below.jpg [ 12.83 KiB | Viewed 1768 times ]
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
Re: The table below provides values of quadratic function f for some value
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23 Feb 2022, 10:47
23 Feb 2022, 10:47
Carcass wrote:
The table below provides values of quadratic function f for some values x. Which of the following equations could define f(x)?
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
Attachment:
GRe The table below.jpg
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
I found this question a little lengthy as I have to pick and choose the numbers to test all equations. Is there a shortcut or strategy to pick no. smartly.
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Re: The table below provides values of quadratic function f for some value
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24 Feb 2022, 05:59
24 Feb 2022, 05:59
1
Carcass wrote:
The table below provides values of quadratic function f for some values x. Which of the following equations could define f(x)?
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
Attachment:
GRe The table below.jpg
(A) \(f(x) = x^2 + x - 1 \)
(B) \(f(x) = x^2 - 1\)
(C) \(f(x) = -x^2 + x + 1 \)
(D) \(f(x) = -x^2 + 1\)
(E) \(f(x) = x^2 - x + 1 \)
Let's eliminate answers choices by plugging easy-to-work-with numbers into each function.
For example, the table tells us that f(0) = 1, so let's plug x = 0 into each function and eliminate those that don't evaluate to 1.
We get:
(A) \(f(0) = 0^2 + 0 - 1 =-1\). Eliminate
(B) \(f(0) = 0^2 - 1 = -1\). Eliminate
(C) \(f(0) = -(0^2) + 0 + 1 = 1\). Keep
(D) \(f(0) = -(0^2) + 1 = 1\). Keep
(E) \(f(0) = 0^2 - 0 + 1 = 1\). Keep
The table also tells us that, when f(1) = 0, so let's plug x = 1 into each remaining function and eliminate those that don't evaluate to 0.
We get:
(C) \(f(1) = -(1^2) + 1 + 1 = 1 \). Eliminate
(D) \(f(1) = -(1^2) + 1 = 0\). Keep
(E) \(f(1) = 1^2 - 1 + 1 = 1\). Eliminate
By the process of elimination, the correct answer is D
