Answer :
To determine which of the given values is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex], we need to substitute each value into the function and see if the result is zero. A zero of the function means that the value makes the function equal to zero. Let's evaluate [tex]\( f(x) \)[/tex] for each of the given potential zeros.
1. Substitute [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ f\left(\frac{1}{3}\right) = 9\left(\frac{1}{3}\right)^2 - 54\left(\frac{1}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \frac{1}{9} - 18 - 19 \][/tex]
[tex]\[ = 1 - 18 - 19 \][/tex]
[tex]\[ = 1 - 37 \][/tex]
[tex]\[ = -36 \quad (\text{not zero}) \][/tex]
2. Substitute [tex]\( x = 3 \frac{1}{3} = \frac{10}{3} \)[/tex]:
[tex]\[ f\left(3 \frac{1}{3}\right) = 9\left(\frac{10}{3}\right)^2 - 54\left(\frac{10}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{100}{9}\right) - 54 \cdot \frac{10}{3} - 19 \][/tex]
[tex]\[ = 100 - 180 - 19 \][/tex]
[tex]\[ = 100 - 199 \][/tex]
[tex]\[ = -99 \quad (\text{not zero}) \][/tex]
3. Substitute [tex]\( x = 6 \frac{1}{3} = \frac{19}{3} \)[/tex]:
[tex]\[ f\left(6 \frac{1}{3}\right) = 9\left(\frac{19}{3}\right)^2 - 54\left(\frac{19}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{361}{9}\right) - 54 \cdot \frac{19}{3} - 19 \][/tex]
[tex]\[ = 361 - 342 - 19 \][/tex]
[tex]\[ = 361 - 361 \][/tex]
[tex]\[ = 0 \quad (\text{zero found}) \][/tex]
4. Substitute [tex]\( x = 9 \frac{1}{3} = \frac{28}{3} \)[/tex]:
[tex]\[ f\left(9 \frac{1}{3}\right) = 9\left(\frac{28}{3}\right)^2 - 54\left(\frac{28}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{784}{9}\right) - 54 \cdot \frac{28}{3} - 19 \][/tex]
[tex]\[ = 784 - 504 - 19 \][/tex]
[tex]\[ = 784 - 523 \][/tex]
[tex]\[ = 261 \quad (\text{not zero}) \][/tex]
From the evaluation, we find that [tex]\( f\left(6 \frac{1}{3}\right) = 0 \)[/tex]. Therefore, [tex]\( x = 6 \frac{1}{3} \)[/tex] is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex].
1. Substitute [tex]\( x = \frac{1}{3} \)[/tex]:
[tex]\[ f\left(\frac{1}{3}\right) = 9\left(\frac{1}{3}\right)^2 - 54\left(\frac{1}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \frac{1}{9} - 18 - 19 \][/tex]
[tex]\[ = 1 - 18 - 19 \][/tex]
[tex]\[ = 1 - 37 \][/tex]
[tex]\[ = -36 \quad (\text{not zero}) \][/tex]
2. Substitute [tex]\( x = 3 \frac{1}{3} = \frac{10}{3} \)[/tex]:
[tex]\[ f\left(3 \frac{1}{3}\right) = 9\left(\frac{10}{3}\right)^2 - 54\left(\frac{10}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{100}{9}\right) - 54 \cdot \frac{10}{3} - 19 \][/tex]
[tex]\[ = 100 - 180 - 19 \][/tex]
[tex]\[ = 100 - 199 \][/tex]
[tex]\[ = -99 \quad (\text{not zero}) \][/tex]
3. Substitute [tex]\( x = 6 \frac{1}{3} = \frac{19}{3} \)[/tex]:
[tex]\[ f\left(6 \frac{1}{3}\right) = 9\left(\frac{19}{3}\right)^2 - 54\left(\frac{19}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{361}{9}\right) - 54 \cdot \frac{19}{3} - 19 \][/tex]
[tex]\[ = 361 - 342 - 19 \][/tex]
[tex]\[ = 361 - 361 \][/tex]
[tex]\[ = 0 \quad (\text{zero found}) \][/tex]
4. Substitute [tex]\( x = 9 \frac{1}{3} = \frac{28}{3} \)[/tex]:
[tex]\[ f\left(9 \frac{1}{3}\right) = 9\left(\frac{28}{3}\right)^2 - 54\left(\frac{28}{3}\right) - 19 \][/tex]
[tex]\[ = 9 \cdot \left(\frac{784}{9}\right) - 54 \cdot \frac{28}{3} - 19 \][/tex]
[tex]\[ = 784 - 504 - 19 \][/tex]
[tex]\[ = 784 - 523 \][/tex]
[tex]\[ = 261 \quad (\text{not zero}) \][/tex]
From the evaluation, we find that [tex]\( f\left(6 \frac{1}{3}\right) = 0 \)[/tex]. Therefore, [tex]\( x = 6 \frac{1}{3} \)[/tex] is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex].