Answer :
To identify the quadratic function having a leading coefficient of 3 and a constant term of -12, we need to review each of the given options. A quadratic function is typically expressed in the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are constants, and [tex]\( a \neq 0 \)[/tex].
Let's analyze each option:
1. [tex]\( f(x) = -12x^2 + 3x + 1 \)[/tex]
- Here, the leading coefficient (the coefficient of [tex]\( x^2 \)[/tex]) is [tex]\(-12\)[/tex].
- This does not match the requirement of having a leading coefficient of 3.
2. [tex]\( f(x) = 3x^2 + 11x - 12 \)[/tex]
- Here, the leading coefficient is [tex]\( 3 \)[/tex], which matches our requirement.
- The constant term is [tex]\(-12\)[/tex], which also matches our requirement.
- This function satisfies both conditions: leading coefficient of 3 and constant term of -12.
3. [tex]\( f(x) = 12x^2 + 3x + 3 \)[/tex]
- Here, the leading coefficient is [tex]\( 12 \)[/tex].
- This does not match the requirement of having a leading coefficient of 3.
4. [tex]\( f(x) = 3x - 12 \)[/tex]
- This is a linear function, not a quadratic function, since it does not have a [tex]\( x^2 \)[/tex] term.
- Therefore, it does not meet the requirement of being a quadratic function.
After evaluating all the options, the only function that meets the criteria of having a leading coefficient of 3 and a constant term of -12 is:
[tex]\[ f(x) = 3x^2 + 11x - 12 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{f(x) = 3x^2 + 11x - 12} \][/tex]
Let's analyze each option:
1. [tex]\( f(x) = -12x^2 + 3x + 1 \)[/tex]
- Here, the leading coefficient (the coefficient of [tex]\( x^2 \)[/tex]) is [tex]\(-12\)[/tex].
- This does not match the requirement of having a leading coefficient of 3.
2. [tex]\( f(x) = 3x^2 + 11x - 12 \)[/tex]
- Here, the leading coefficient is [tex]\( 3 \)[/tex], which matches our requirement.
- The constant term is [tex]\(-12\)[/tex], which also matches our requirement.
- This function satisfies both conditions: leading coefficient of 3 and constant term of -12.
3. [tex]\( f(x) = 12x^2 + 3x + 3 \)[/tex]
- Here, the leading coefficient is [tex]\( 12 \)[/tex].
- This does not match the requirement of having a leading coefficient of 3.
4. [tex]\( f(x) = 3x - 12 \)[/tex]
- This is a linear function, not a quadratic function, since it does not have a [tex]\( x^2 \)[/tex] term.
- Therefore, it does not meet the requirement of being a quadratic function.
After evaluating all the options, the only function that meets the criteria of having a leading coefficient of 3 and a constant term of -12 is:
[tex]\[ f(x) = 3x^2 + 11x - 12 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{f(x) = 3x^2 + 11x - 12} \][/tex]
