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Which is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 192 \)[/tex]?

A. [tex]\( x = \frac{1}{3} \)[/tex]
B. [tex]\( x = 3 \frac{1}{3} \)[/tex]
C. [tex]\( x = 6 \frac{1}{3} \)[/tex]
D. [tex]\( x = 9 \frac{1}{3} \)[/tex]

Sagot :

To determine which values are zeroes of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 192 \)[/tex], we will evaluate the function at each of the given points and check if the result is zero. Let's go through each candidate one by one:

1. For [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ f\left(\frac{1}{3}\right) = 9\left(\frac{1}{3}\right)^2 - 54\left(\frac{1}{3}\right) - 192 \][/tex]
Simplify inside the parentheses:
[tex]\[ f\left(\frac{1}{3}\right) = 9 \left(\frac{1}{3} \times \frac{1}{3}\right) - 54 \times \frac{1}{3} - 192 \][/tex]
[tex]\[ = 9 \times \frac{1}{9} - 18 - 192 \][/tex]
[tex]\[ = 1 - 18 - 192 \][/tex]
[tex]\[ = -209 \][/tex]
Thus, [tex]\( f\left(\frac{1}{3}\right) \neq 0 \)[/tex].

2. For [tex]\( x = 3 \frac{1}{3} = \frac{10}{3} \)[/tex]:
[tex]\[ f\left(\frac{10}{3}\right) = 9\left(\frac{10}{3}\right)^2 - 54\left(\frac{10}{3}\right) - 192 \][/tex]
Simplify inside the parentheses:
[tex]\[ f\left(\frac{10}{3}\right) = 9 \left(\frac{100}{9}\right) - 54 \times \frac{10}{3} - 192 \][/tex]
[tex]\[ = 9 \times \frac{100}{9} - 180 - 192 \][/tex]
[tex]\[ = 100 - 180 - 192 \][/tex]
[tex]\[ = -272 \][/tex]
Thus, [tex]\( f\left(\frac{10}{3}\right) \neq 0 \)[/tex].

3. For [tex]\( x = 6 \frac{1}{3} = \frac{19}{3} \)[/tex]:
[tex]\[ f\left(\frac{19}{3}\right) = 9\left(\frac{19}{3}\right)^2 - 54\left(\frac{19}{3}\right) - 192 \][/tex]
Simplify inside the parentheses:
[tex]\[ f\left(\frac{19}{3}\right) = 9 \left(\frac{361}{9}\right) - 54 \times \frac{19}{3} - 192 \][/tex]
[tex]\[ = 9 \times \frac{361}{9} - 342 - 192 \][/tex]
[tex]\[ = 361 - 342 - 192 \][/tex]
[tex]\[ = -173 \][/tex]
Thus, [tex]\( f\left(\frac{19}{3}\right) \neq 0 \)[/tex].

4. For [tex]\( x = 9 \frac{1}{3} = \frac{28}{3} \)[/tex]:
[tex]\[ f\left(\frac{28}{3}\right) = 9\left(\frac{28}{3}\right)^2 - 54\left(\frac{28}{3}\right) - 192 \][/tex]
Simplify inside the parentheses:
[tex]\[ f\left(\frac{28}{3}\right) = 9 \left(\frac{784}{9}\right) - 54 \times \frac{28}{3} - 192 \][/tex]
[tex]\[ = 9 \times \frac{784}{9} - 504 - 192 \][/tex]
[tex]\[ = 784 - 504 - 192 \][/tex]
[tex]\[ = 88 \][/tex]
Thus, [tex]\( f\left(\frac{28}{3}\right) \neq 0 \)[/tex].

Since none of the given values satisfy [tex]\( f(x) = 0 \)[/tex], we conclude that none of them is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 192 \)[/tex]. Therefore, the answer is:
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