Question
find the range of the function f, of, x, equals, x, squared, minus, 3, .f(x)=x
2
−3.
2
−3.
Answers
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Answered by
GPT-4o mini
To find the range of the function \( f(x) = x^2 - 3 \), we can analyze its structure.
1. **Identify the type of function**: The function \( f(x) = x^2 - 3 \) is a quadratic function.
2. **Find the vertex**: The quadratic function \( f(x) = x^2 - 3 \) opens upwards (since the coefficient of \( x^2 \) is positive). The vertex of the function, which represents its minimum value, occurs at \( x = 0 \) (the vertex formula for \( f(x) = ax^2 + bx + c \) gives the vertex at \( x = -\frac{b}{2a} \), here \( b = 0 \) and \( a = 1 \)).
Substituting \( x = 0 \) into the function:
\[
f(0) = 0^2 - 3 = -3
\]
3. **Determine the range**: Since the vertex represents the minimum value and the function opens upwards, the function can take any value greater than or equal to this minimum value.
Thus, the range of the function \( f(x) = x^2 - 3 \) is:
\[
\text{Range}(f) = [-3, \infty)
\]
1. **Identify the type of function**: The function \( f(x) = x^2 - 3 \) is a quadratic function.
2. **Find the vertex**: The quadratic function \( f(x) = x^2 - 3 \) opens upwards (since the coefficient of \( x^2 \) is positive). The vertex of the function, which represents its minimum value, occurs at \( x = 0 \) (the vertex formula for \( f(x) = ax^2 + bx + c \) gives the vertex at \( x = -\frac{b}{2a} \), here \( b = 0 \) and \( a = 1 \)).
Substituting \( x = 0 \) into the function:
\[
f(0) = 0^2 - 3 = -3
\]
3. **Determine the range**: Since the vertex represents the minimum value and the function opens upwards, the function can take any value greater than or equal to this minimum value.
Thus, the range of the function \( f(x) = x^2 - 3 \) is:
\[
\text{Range}(f) = [-3, \infty)
\]
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