To determine which of the given values is a zero of the quadratic function [tex]\( f(x) = 16x^2 + 32x \)[/tex], we need to evaluate the function at each of the given potential zeros and see if the result is zero. A zero of the function [tex]\( f(x) \)[/tex] satisfies [tex]\( f(x) = 0 \)[/tex].
The given potential zeros are:
- [tex]\( x = -5.25 \)[/tex]
- [tex]\( x = -2.25 \)[/tex]
- [tex]\( x = -1.25 \)[/tex]
- [tex]\( x = -0.25 \)[/tex]
Let's evaluate each potential zero.
1. Evaluate [tex]\( f(-5.25) \)[/tex]:
[tex]\[ f(-5.25) = 16(-5.25)^2 + 32(-5.25) \][/tex]
Calculating:
[tex]\[ (-5.25)^2 = 27.5625 \][/tex]
[tex]\[ 16 \cdot 27.5625 = 441 \][/tex]
[tex]\[ 32 \cdot (-5.25) = -168 \][/tex]
[tex]\[ f(-5.25) = 441 - 168 = 273 \][/tex]
Since [tex]\( f(-5.25) \neq 0 \)[/tex], [tex]\( x = -5.25 \)[/tex] is not a zero of the function.
2. Evaluate [tex]\( f(-2.25) \)[/tex]:
[tex]\[ f(-2.25) = 16(-2.25)^2 + 32(-2.25) \][/tex]
Calculating:
[tex]\[ (-2.25)^2 = 5.0625 \][/tex]
[tex]\[ 16 \cdot 5.0625 = 81 \][/tex]
[tex]\[ 32 \cdot (-2.25) = -72 \][/tex]
[tex]\[ f(-2.25) = 81 - 72 = 9 \][/tex]
Since [tex]\( f(-2.25) \neq 0 \)[/tex], [tex]\( x = -2.25 \)[/tex] is not a zero of the function.
3. Evaluate [tex]\( f(-1.25) \)[/tex]:
[tex]\[ f(-1.25) = 16(-1.25)^2 + 32(-1.25) \][/tex]
Calculating:
[tex]\[ (-1.25)^2 = 1.5625 \][/tex]
[tex]\[ 16 \cdot 1.5625 = 25 \][/tex]
[tex]\[ 32 \cdot (-1.25) = -40 \][/tex]
[tex]\[ f(-1.25) = 25 - 40 = -15 \][/tex]
Since [tex]\( f(-1.25) \neq 0 \)[/tex], [tex]\( x = -1.25 \)[/tex] is not a zero of the function.
4. Evaluate [tex]\( f(-0.25) \)[/tex]:
[tex]\[ f(-0.25) = 16(-0.25)^2 + 32(-0.25) \][/tex]
Calculating:
[tex]\[ (-0.25)^2 = 0.0625 \][/tex]
[tex]\[ 16 \cdot 0.0625 = 1 \][/tex]
[tex]\[ 32 \cdot (-0.25) = -8 \][/tex]
[tex]\[ f(-0.25) = 1 - 8 = -7 \][/tex]
Since [tex]\( f(-0.25) \neq 0 \)[/tex], [tex]\( x = -0.25 \)[/tex] is not a zero of the function.
After evaluating the function at each potential zero, we find that none of the given values are zeros of the function [tex]\( f(x) = 16x^2 + 32x \)[/tex].
