Answer :
To determine the vertex of the quadratic function [tex]\( f(x) \)[/tex], let's examine each given form of the function.
Form (A): [tex]\( f(x)=3(x+6)^2-75 \)[/tex]
This form of the quadratic function is known as the vertex form. The vertex form of a quadratic function is given by:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Given the function [tex]\( f(x)=3(x+6)^2-75 \)[/tex]:
- We can identify [tex]\( (x + 6) \)[/tex] as [tex]\( (x - (-6)) \)[/tex], which means [tex]\( h = -6 \)[/tex].
- The constant term is [tex]\(-75\)[/tex], so [tex]\( k = -75 \)[/tex].
Thus, the vertex form of the function quickly reveals that the vertex is [tex]\((-6, -75)\)[/tex].
Form (B): [tex]\( f(x)=3x^2+36x+33 \)[/tex]
This form is the standard form of the quadratic function, [tex]\( f(x) = ax^2 + bx + c \)[/tex].
To find the vertex from the standard form, we typically use the vertex formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex] and [tex]\( b = 36 \)[/tex].
Using the formula:
[tex]\[ h = -\frac{36}{2 \cdot 3} = -\frac{36}{6} = -6 \][/tex]
Next, we substitute [tex]\( h = -6 \)[/tex] back into the function to find [tex]\( k \)[/tex]:
[tex]\[ f(-6) = 3(-6)^2 + 36(-6) + 33 = 3(36) - 216 + 33 = 108 - 216 + 33 = -75 \][/tex]
So, the vertex is [tex]\((-6, -75)\)[/tex].
Form (C): [tex]\( f(x)=3(x+1)(x+11) \)[/tex]
This form is the factored form of the quadratic function. To find the vertex, we need to find the midpoint of the roots:
The roots of the function occur where each factor is zero:
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
[tex]\[ x + 11 = 0 \implies x = -11 \][/tex]
The average of the roots gives the x-coordinate of the vertex:
[tex]\[ h = \frac{-1 + (-11)}{2} = \frac{-12}{2} = -6 \][/tex]
Substitute [tex]\( h = -6 \)[/tex] back into the function to find [tex]\( k \)[/tex]:
[tex]\[ f(-6) = 3(-6 + 1)(-6 + 11) = 3(-5)(5) = 3(-25) = -75 \][/tex]
So, the vertex is [tex]\((-6, -75)\)[/tex].
Conclusion:
The vertex form (A), [tex]\( f(x)=3(x+6)^2-75 \)[/tex], most quickly reveals the vertex because it is directly readable from the equation.
The vertex is [tex]\( \boxed{(-6, -75)} \)[/tex].
Form (A): [tex]\( f(x)=3(x+6)^2-75 \)[/tex]
This form of the quadratic function is known as the vertex form. The vertex form of a quadratic function is given by:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Given the function [tex]\( f(x)=3(x+6)^2-75 \)[/tex]:
- We can identify [tex]\( (x + 6) \)[/tex] as [tex]\( (x - (-6)) \)[/tex], which means [tex]\( h = -6 \)[/tex].
- The constant term is [tex]\(-75\)[/tex], so [tex]\( k = -75 \)[/tex].
Thus, the vertex form of the function quickly reveals that the vertex is [tex]\((-6, -75)\)[/tex].
Form (B): [tex]\( f(x)=3x^2+36x+33 \)[/tex]
This form is the standard form of the quadratic function, [tex]\( f(x) = ax^2 + bx + c \)[/tex].
To find the vertex from the standard form, we typically use the vertex formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
where [tex]\( a = 3 \)[/tex] and [tex]\( b = 36 \)[/tex].
Using the formula:
[tex]\[ h = -\frac{36}{2 \cdot 3} = -\frac{36}{6} = -6 \][/tex]
Next, we substitute [tex]\( h = -6 \)[/tex] back into the function to find [tex]\( k \)[/tex]:
[tex]\[ f(-6) = 3(-6)^2 + 36(-6) + 33 = 3(36) - 216 + 33 = 108 - 216 + 33 = -75 \][/tex]
So, the vertex is [tex]\((-6, -75)\)[/tex].
Form (C): [tex]\( f(x)=3(x+1)(x+11) \)[/tex]
This form is the factored form of the quadratic function. To find the vertex, we need to find the midpoint of the roots:
The roots of the function occur where each factor is zero:
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
[tex]\[ x + 11 = 0 \implies x = -11 \][/tex]
The average of the roots gives the x-coordinate of the vertex:
[tex]\[ h = \frac{-1 + (-11)}{2} = \frac{-12}{2} = -6 \][/tex]
Substitute [tex]\( h = -6 \)[/tex] back into the function to find [tex]\( k \)[/tex]:
[tex]\[ f(-6) = 3(-6 + 1)(-6 + 11) = 3(-5)(5) = 3(-25) = -75 \][/tex]
So, the vertex is [tex]\((-6, -75)\)[/tex].
Conclusion:
The vertex form (A), [tex]\( f(x)=3(x+6)^2-75 \)[/tex], most quickly reveals the vertex because it is directly readable from the equation.
The vertex is [tex]\( \boxed{(-6, -75)} \)[/tex].