Question
The graph of a function f(x)
passes through the following points:
(0,2), (1,0), (-1,0)
Which of the following could be f(x)
?
passes through the following points:
(0,2), (1,0), (-1,0)
Which of the following could be f(x)
?
Answers
Answered by
GPT-4o mini
To determine possible forms of the function \( f(x) \) that passes through the given points \((0, 2)\), \((1, 0)\), and \((-1, 0)\), we take the following into consideration:
1. The point \((0, 2)\) indicates that when \( x = 0 \), \( f(x) = 2\).
2. The point \((1, 0)\) indicates that \( f(1) = 0\) which means that \( x = 1\) is a root of the function.
3. The point \((-1, 0)\) indicates that \( f(-1) = 0\) which means that \( x = -1\) is another root of the function.
Since \( x = 1\) and \( x = -1\) are roots, we can express \( f(x) \) in a factored form. A simple polynomial that satisfies these conditions could take the form of:
\[
f(x) = k(x - 1)(x + 1) + 2
\]
where \( k \) is a constant that would help determine the vertical scaling of the parabola.
Expanding this gives:
\[
f(x) = k(x^2 - 1) + 2 = kx^2 - k + 2
\]
Next, we can use the point \((0, 2)\) to find \( k \):
\[
f(0) = k(0)^2 - k + 2 = 2
\]
\[
-k + 2 = 2 \implies -k = 0 \implies k = 0
\]
Since \( k = 0\) doesn’t create a function with roots, we need to include a non-zero value for \( k \).
If we consider \( k\) is any non-zero constant, we can explore logarithmic and polynomial combinations. Alternatively, using the condition that \( f(1) = 0\) and \( f(-1) = 0\) again gives roots to the polynomial. Another potential helper could be:
\[
f(x) = a(x^2 - 1) + 2
\]
Where \( a \) is adjusted to maintain the y-intercept.
Assuming \( a > 0\) and non-constant formulations involve sine, although linear combinations would also need higher-degree polynomials to maintain these roots.
Thus, viable candidates that meet these criteria include quadratic polynomial forms that approach stable Y-values while intersecting the axis, such as:
\[
f(x) = ax^2 - 1 + 2
\]
\[
= ax^2 + 1
\]
To finalize without specific candidates required, check against any polynomial (linear, quadratic with above conditions lead) or enabling constructive combinations that intersect \( (0,2) \) while maintaining roots with each set.
1. The point \((0, 2)\) indicates that when \( x = 0 \), \( f(x) = 2\).
2. The point \((1, 0)\) indicates that \( f(1) = 0\) which means that \( x = 1\) is a root of the function.
3. The point \((-1, 0)\) indicates that \( f(-1) = 0\) which means that \( x = -1\) is another root of the function.
Since \( x = 1\) and \( x = -1\) are roots, we can express \( f(x) \) in a factored form. A simple polynomial that satisfies these conditions could take the form of:
\[
f(x) = k(x - 1)(x + 1) + 2
\]
where \( k \) is a constant that would help determine the vertical scaling of the parabola.
Expanding this gives:
\[
f(x) = k(x^2 - 1) + 2 = kx^2 - k + 2
\]
Next, we can use the point \((0, 2)\) to find \( k \):
\[
f(0) = k(0)^2 - k + 2 = 2
\]
\[
-k + 2 = 2 \implies -k = 0 \implies k = 0
\]
Since \( k = 0\) doesn’t create a function with roots, we need to include a non-zero value for \( k \).
If we consider \( k\) is any non-zero constant, we can explore logarithmic and polynomial combinations. Alternatively, using the condition that \( f(1) = 0\) and \( f(-1) = 0\) again gives roots to the polynomial. Another potential helper could be:
\[
f(x) = a(x^2 - 1) + 2
\]
Where \( a \) is adjusted to maintain the y-intercept.
Assuming \( a > 0\) and non-constant formulations involve sine, although linear combinations would also need higher-degree polynomials to maintain these roots.
Thus, viable candidates that meet these criteria include quadratic polynomial forms that approach stable Y-values while intersecting the axis, such as:
\[
f(x) = ax^2 - 1 + 2
\]
\[
= ax^2 + 1
\]
To finalize without specific candidates required, check against any polynomial (linear, quadratic with above conditions lead) or enabling constructive combinations that intersect \( (0,2) \) while maintaining roots with each set.
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