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Sagot :
To solve this problem, we first need to identify the quadratic function based on the given points [tex]\((0, 0)\)[/tex], [tex]\((1, 3)\)[/tex], [tex]\((2, 12)\)[/tex], and [tex]\((3, 27)\)[/tex].
1. Formulating the Quadratic Function:
The general form of a quadratic function is [tex]\( y = ax^2 + bx + c \)[/tex].
- From the given point [tex]\((0, 0)\)[/tex], we can determine that [tex]\( c = 0 \)[/tex] because when [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex].
2. Using Points to Find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
We will now use the points [tex]\((1, 3)\)[/tex] and [tex]\((2, 12)\)[/tex] to form a system of equations:
- For point [tex]\((1, 3)\)[/tex]:
[tex]\[ 3 = a(1)^2 + b(1) + 0 \implies a + b = 3 \][/tex]
- For point [tex]\((2, 12)\)[/tex]:
[tex]\[ 12 = a(2)^2 + b(2) + 0 \implies 4a + 2b = 12 \][/tex]
3. Solving the System of Equations:
We have the system:
[tex]\[ \begin{cases} a + b = 3 \\ 4a + 2b = 12 \end{cases} \][/tex]
- Multiply the first equation by 2:
[tex]\[ 2a + 2b = 6 \][/tex]
- Subtract this from the second equation:
[tex]\[ (4a + 2b) - (2a + 2b) = 12 - 6 \implies 2a = 6 \implies a = 3 \][/tex]
- Substitute [tex]\( a = 3 \)[/tex] back into [tex]\( a + b = 3 \)[/tex]:
[tex]\[ 3 + b = 3 \implies b = 0 \][/tex]
Therefore, the quadratic function is:
[tex]\[ y = 3x^2 \][/tex]
4. Comparing with an Exponential Function:
To find an exponential function that grows faster than [tex]\( y = 3x^2 \)[/tex] in the given range [tex]\( 0 < x < 3 \)[/tex], we consider exponential functions of the form [tex]\( y = e^x \)[/tex] or [tex]\( y = 2^x \)[/tex].
5. Evaluating Growth Rates:
- For [tex]\( y = 2^x \)[/tex] at [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2^3 = 8 \][/tex]
- For [tex]\( y = 3x^2 \)[/tex] at [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 \cdot 3^2 = 3 \cdot 9 = 27 \][/tex]
Since the quadratic function [tex]\( y = 3x^2 \)[/tex] at [tex]\( x = 3 \)[/tex] gives [tex]\( y = 27 \)[/tex], which exceeds the growth of the exponential function [tex]\( y = 2^x \)[/tex] at [tex]\( x = 3 \)[/tex], we can infer that for larger bases in the exponential form, the exponential function will eventually grow faster.
Thus, [tex]\( y = e^x \)[/tex] or higher bases like [tex]\( y = 3^x \)[/tex] and [tex]\( y = 4^x \)[/tex] grow faster than the given quadratic function as [tex]\( x \)[/tex] increases, but in the range [tex]\( 0 < x < 3 \)[/tex], identifying even moderate exponential growth, [tex]\( y = 2^x \)[/tex] already approaches the growth rate significantly though does not exceed at x=3 specifically.
Conclusively, functions such as [tex]\( y = e^x \)[/tex] or [tex]\( y = 3^x \)[/tex] ultimately show faster growth compared to [tex]\( y = 3x^2 \)[/tex] ultimately at larger [tex]\( x \)[/tex], beyond narrow intervals.
1. Formulating the Quadratic Function:
The general form of a quadratic function is [tex]\( y = ax^2 + bx + c \)[/tex].
- From the given point [tex]\((0, 0)\)[/tex], we can determine that [tex]\( c = 0 \)[/tex] because when [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex].
2. Using Points to Find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
We will now use the points [tex]\((1, 3)\)[/tex] and [tex]\((2, 12)\)[/tex] to form a system of equations:
- For point [tex]\((1, 3)\)[/tex]:
[tex]\[ 3 = a(1)^2 + b(1) + 0 \implies a + b = 3 \][/tex]
- For point [tex]\((2, 12)\)[/tex]:
[tex]\[ 12 = a(2)^2 + b(2) + 0 \implies 4a + 2b = 12 \][/tex]
3. Solving the System of Equations:
We have the system:
[tex]\[ \begin{cases} a + b = 3 \\ 4a + 2b = 12 \end{cases} \][/tex]
- Multiply the first equation by 2:
[tex]\[ 2a + 2b = 6 \][/tex]
- Subtract this from the second equation:
[tex]\[ (4a + 2b) - (2a + 2b) = 12 - 6 \implies 2a = 6 \implies a = 3 \][/tex]
- Substitute [tex]\( a = 3 \)[/tex] back into [tex]\( a + b = 3 \)[/tex]:
[tex]\[ 3 + b = 3 \implies b = 0 \][/tex]
Therefore, the quadratic function is:
[tex]\[ y = 3x^2 \][/tex]
4. Comparing with an Exponential Function:
To find an exponential function that grows faster than [tex]\( y = 3x^2 \)[/tex] in the given range [tex]\( 0 < x < 3 \)[/tex], we consider exponential functions of the form [tex]\( y = e^x \)[/tex] or [tex]\( y = 2^x \)[/tex].
5. Evaluating Growth Rates:
- For [tex]\( y = 2^x \)[/tex] at [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2^3 = 8 \][/tex]
- For [tex]\( y = 3x^2 \)[/tex] at [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 3 \cdot 3^2 = 3 \cdot 9 = 27 \][/tex]
Since the quadratic function [tex]\( y = 3x^2 \)[/tex] at [tex]\( x = 3 \)[/tex] gives [tex]\( y = 27 \)[/tex], which exceeds the growth of the exponential function [tex]\( y = 2^x \)[/tex] at [tex]\( x = 3 \)[/tex], we can infer that for larger bases in the exponential form, the exponential function will eventually grow faster.
Thus, [tex]\( y = e^x \)[/tex] or higher bases like [tex]\( y = 3^x \)[/tex] and [tex]\( y = 4^x \)[/tex] grow faster than the given quadratic function as [tex]\( x \)[/tex] increases, but in the range [tex]\( 0 < x < 3 \)[/tex], identifying even moderate exponential growth, [tex]\( y = 2^x \)[/tex] already approaches the growth rate significantly though does not exceed at x=3 specifically.
Conclusively, functions such as [tex]\( y = e^x \)[/tex] or [tex]\( y = 3^x \)[/tex] ultimately show faster growth compared to [tex]\( y = 3x^2 \)[/tex] ultimately at larger [tex]\( x \)[/tex], beyond narrow intervals.
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