To determine which form of the quadratic function displays the zeros of the function [tex]\( h \)[/tex], let's analyze each form and look for the one that clearly shows the zeros (the values of [tex]\( x \)[/tex] for which [tex]\( h(x) = 0 \)[/tex]).
### Form A: [tex]\( h(x) = -4x^2 + 16 \)[/tex]
This is a standard quadratic form written as [tex]\( h(x) = ax^2 + bx + c \)[/tex]. However, finding the zeros directly from this form requires solving the equation [tex]\( -4x^2 + 16 = 0 \)[/tex], which involves some steps of algebraic manipulation.
### Form B: [tex]\( h(x) = -4(x-2)(x+2) \)[/tex]
This form is a factored form of the quadratic equation. In this form, the zeros are readily identifiable because [tex]\( h(x) \)[/tex] equals zero when either factor is zero:
- [tex]\((x-2) = 0 \)[/tex] gives [tex]\( x = 2 \)[/tex]
- [tex]\((x+2) = 0 \)[/tex] gives [tex]\( x = -2 \)[/tex]
Hence, the zeros are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex] which are clear from the form itself.
### Form C: [tex]\( h(x) = -4(x^2 - 4) \)[/tex]
This form can be recognized as a variation of a quadratic function. However, to find the zeros, we need to solve [tex]\( x^2 - 4 = 0 \)[/tex] and then divide by -4, still an extra step is required.
### Form D: [tex]\( h(x) = -2(2x^2 - 8) \)[/tex]
Similar to Form C, this form shows the quadratic function but in a slightly transformed way. To find the zeros directly is not immediately clear and requires extra steps to solve the equation.
### Conclusion
Given the choices, the form that clearly displays the zeros of the function [tex]\( h \)[/tex] is:
[tex]\[ \text{B. } h(x) = -4(x-2)(x+2) \][/tex]
This form immediately shows that the zeros of the function are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
