Which is a zero of the quadratic function [tex]f(x) = 16x^2 + 32x - 9[/tex]?

A. [tex]x = -5.25[/tex]
B. [tex]x = -2.25[/tex]
C. [tex]x = -1.25[/tex]
D. [tex]x = -0.25[/tex]



Answer :

To determine which value is a zero of the quadratic function [tex]\( f(x) = 16x^2 + 32x - 9 \)[/tex], we need to evaluate the function at each of the given possible zeros and see if it equals zero.

Let’s evaluate the function step-by-step for each given value:

1. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -5.25 \)[/tex]

[tex]\[ f(-5.25) = 16(-5.25)^2 + 32(-5.25) - 9 \][/tex]
Calculate [tex]\( (-5.25)^2 \)[/tex]:
[tex]\[ (-5.25)^2 = 27.5625 \][/tex]
Substitute back into the function:
[tex]\[ f(-5.25) = 16(27.5625) + 32(-5.25) - 9 \][/tex]
Calculate each term:
[tex]\[ 16 \cdot 27.5625 = 441 \][/tex]
[tex]\[ 32 \cdot (-5.25) = -168 \][/tex]
Add the terms together:
[tex]\[ f(-5.25) = 441 - 168 - 9 = 264 \][/tex]
Since [tex]\( f(-5.25) \neq 0 \)[/tex], [tex]\( x = -5.25 \)[/tex] is not a zero of the function.

2. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -2.25 \)[/tex]

[tex]\[ f(-2.25) = 16(-2.25)^2 + 32(-2.25) - 9 \][/tex]
Calculate [tex]\( (-2.25)^2 \)[/tex]:
[tex]\[ (-2.25)^2 = 5.0625 \][/tex]
Substitute back into the function:
[tex]\[ f(-2.25) = 16(5.0625) + 32(-2.25) - 9 \][/tex]
Calculate each term:
[tex]\[ 16 \cdot 5.0625 = 81 \][/tex]
[tex]\[ 32 \cdot (-2.25) = -72 \][/tex]
Add the terms together:
[tex]\[ f(-2.25) = 81 - 72 - 9 = 0 \][/tex]
Since [tex]\( f(-2.25) = 0 \)[/tex], [tex]\( x = -2.25 \)[/tex] is indeed a zero of the function.

3. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -125 \)[/tex]

[tex]\[ f(-125) = 16(-125)^2 + 32(-125) - 9 \][/tex]
Calculate [tex]\( (-125)^2 \)[/tex]:
[tex]\[ (-125)^2 = 15625 \][/tex]
Substitute back into the function:
[tex]\[ f(-125) = 16(15625) + 32(-125) - 9 \][/tex]
Calculate each term:
[tex]\[ 16 \cdot 15625 = 250000 \][/tex]
[tex]\[ 32 \cdot (-125) = -4000 \][/tex]
Add the terms together:
[tex]\[ f(-125) = 250000 - 4000 - 9 = 245991 \][/tex]
Since [tex]\( f(-125) \neq 0 \)[/tex], [tex]\( x = -125 \)[/tex] is not a zero of the function.

4. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -0.25 \)[/tex]

[tex]\[ f(-0.25) = 16(-0.25)^2 + 32(-0.25) - 9 \][/tex]
Calculate [tex]\( (-0.25)^2 \)[/tex]:
[tex]\[ (-0.25)^2 = 0.0625 \][/tex]
Substitute back into the function:
[tex]\[ f(-0.25) = 16(0.0625) + 32(-0.25) - 9 \][/tex]
Calculate each term:
[tex]\[ 16 \cdot 0.0625 = 1 \][/tex]
[tex]\[ 32 \cdot (-0.25) = -8 \][/tex]
Add the terms together:
[tex]\[ f(-0.25) = 1 - 8 - 9 = -16 \][/tex]
Since [tex]\( f(-0.25) \neq 0 \)[/tex], [tex]\( x = -0.25 \)[/tex] is not a zero of the function.

From all the evaluations, the value that makes the function equal to zero is [tex]\( x = -2.25 \)[/tex]. Thus, the correct answer is:

[tex]\( x = -2.25 \)[/tex]