Answer :
To determine which form of the given quadratic function [tex]\( h \)[/tex] displays the zeros of the function, let's first understand the concept of zeros (roots) of a quadratic function. The zeros of a quadratic function are the values of [tex]\( x \)[/tex] for which the function [tex]\( h(x) = 0 \)[/tex].
The factored form of a quadratic function is particularly useful for identifying the zeros because it expresses the function as a product of linear factors. A general factored form of a quadratic function is:
[tex]\[ h(x) = a(x - r_1)(x - r_2) \][/tex]
where [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the zeros of the function.
Now, let's analyze each option:
### Option A
[tex]\[ h(x) = -4x^2 + 16 \][/tex]
This is in standard form. It is not immediately clear what the zeros are from this representation.
### Option B
[tex]\[ h(x) = -4(x-2)(x+2) \][/tex]
This is in factored form, and it clearly shows the zeros of the function. We can see that the function will be zero when:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \][/tex]
Thus, the zeros are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### Option C
[tex]\[ h(x) = -4(x^2 - 4) \][/tex]
This is in a form that is very close to being factored, but not entirely. To find the zeros, we must further factor [tex]\( x^2 - 4 \)[/tex] into [tex]\( (x - 2)(x + 2) \)[/tex], leading us to the same zeros as in option B. However, this intermediate step is not shown explicitly here.
### Option D
[tex]\[ h(x) = -2(2x^2 - 8) \][/tex]
This is again a form that is close to factoring but isn't fully factored. To identify the zeros directly, further factoring would be required.
Given all of this analysis, the form that most clearly shows the zeros of the function is option B.
[tex]\[ \boxed{h(x) = -4(x-2)(x+2)} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
The factored form of a quadratic function is particularly useful for identifying the zeros because it expresses the function as a product of linear factors. A general factored form of a quadratic function is:
[tex]\[ h(x) = a(x - r_1)(x - r_2) \][/tex]
where [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the zeros of the function.
Now, let's analyze each option:
### Option A
[tex]\[ h(x) = -4x^2 + 16 \][/tex]
This is in standard form. It is not immediately clear what the zeros are from this representation.
### Option B
[tex]\[ h(x) = -4(x-2)(x+2) \][/tex]
This is in factored form, and it clearly shows the zeros of the function. We can see that the function will be zero when:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \][/tex]
Thus, the zeros are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
### Option C
[tex]\[ h(x) = -4(x^2 - 4) \][/tex]
This is in a form that is very close to being factored, but not entirely. To find the zeros, we must further factor [tex]\( x^2 - 4 \)[/tex] into [tex]\( (x - 2)(x + 2) \)[/tex], leading us to the same zeros as in option B. However, this intermediate step is not shown explicitly here.
### Option D
[tex]\[ h(x) = -2(2x^2 - 8) \][/tex]
This is again a form that is close to factoring but isn't fully factored. To identify the zeros directly, further factoring would be required.
Given all of this analysis, the form that most clearly shows the zeros of the function is option B.
[tex]\[ \boxed{h(x) = -4(x-2)(x+2)} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{B} \][/tex]