Answer :
To determine which, if any, of the given values [tex]\( x = \frac{1}{3} \)[/tex], [tex]\( x = 3 \frac{1}{3} \)[/tex], [tex]\( x = 6 \frac{1}{3} \)[/tex], and [tex]\( x = 9 \frac{1}{3} \)[/tex] are zeros of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex], we need to evaluate the function at each of these points and check if the value is zero.
The points to evaluate are:
1. [tex]\( x = \frac{1}{3} \)[/tex]
2. [tex]\( x = 3 \frac{1}{3} \)[/tex]
3. [tex]\( x = 6 \frac{1}{3} \)[/tex]
4. [tex]\( x = 9 \frac{1}{3} \)[/tex]
Let's evaluate the function [tex]\( f(x) \)[/tex] at these points.
1. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = \frac{1}{3} \)[/tex]:
- Plug [tex]\( x = \frac{1}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{1}{3}\right) = 9\left(\frac{1}{3}\right)^2 - 54\left(\frac{1}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{1}{3}\right) = 9 \cdot \frac{1}{9} - 54 \cdot \frac{1}{3} - 19 = 1 - 18 - 19 = -36 \][/tex]
Hence, [tex]\( f\left(\frac{1}{3}\right) = -36 \)[/tex].
2. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 3 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 3 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{10}{3} \)[/tex].
- Plug [tex]\( x = \frac{10}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{10}{3}\right) = 9\left(\frac{10}{3}\right)^2 - 54\left(\frac{10}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{10}{3}\right) = 9 \cdot \left(\frac{100}{9}\right) - 54 \cdot \left(\frac{10}{3}\right) - 19 = 100 - 180 - 19 = -99 \][/tex]
Hence, [tex]\( f\left(\frac{10}{3}\right) = -99 \)[/tex].
3. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 6 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 6 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{19}{3} \)[/tex].
- Plug [tex]\( x = \frac{19}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{19}{3}\right) = 9\left(\frac{19}{3}\right)^2 - 54\left(\frac{19}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{19}{3}\right) = 9 \cdot \left(\frac{361}{9}\right) - 54 \cdot \left(\frac{19}{3}\right) - 19 = 361 - 342 - 19 \approx 0 \][/tex]
Hence, [tex]\( f\left(\frac{19}{3}\right) \approx 0 \)[/tex].
4. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 9 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 9 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{28}{3} \)[/tex].
- Plug [tex]\( x = \frac{28}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{28}{3}\right) = 9\left(\frac{28}{3}\right)^2 - 54\left(\frac{28}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{28}{3}\right) = 9 \cdot \left(\frac{784}{9}\right) - 54 \cdot \left(\frac{28}{3}\right) - 19 = 784 - 504 - 19 = 261 \][/tex]
Hence, [tex]\( f\left(\frac{28}{3}\right) = 261 \)[/tex].
By evaluating the function [tex]\( f(x) \)[/tex] at these points, we observe that:
- [tex]\( f\left(\frac{1}{3}\right) = -36 \)[/tex]
- [tex]\( f\left(\frac{10}{3}\right) = -99 \)[/tex]
- [tex]\( f\left(\frac{19}{3}\right) \approx 0 \)[/tex]
- [tex]\( f\left(\frac{28}{3}\right) = 261 \)[/tex]
Therefore, the value [tex]\( x = 6 \frac{1}{3} \)[/tex] (or [tex]\( x = \frac{19}{3} \)[/tex]) is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex].
The points to evaluate are:
1. [tex]\( x = \frac{1}{3} \)[/tex]
2. [tex]\( x = 3 \frac{1}{3} \)[/tex]
3. [tex]\( x = 6 \frac{1}{3} \)[/tex]
4. [tex]\( x = 9 \frac{1}{3} \)[/tex]
Let's evaluate the function [tex]\( f(x) \)[/tex] at these points.
1. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = \frac{1}{3} \)[/tex]:
- Plug [tex]\( x = \frac{1}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{1}{3}\right) = 9\left(\frac{1}{3}\right)^2 - 54\left(\frac{1}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{1}{3}\right) = 9 \cdot \frac{1}{9} - 54 \cdot \frac{1}{3} - 19 = 1 - 18 - 19 = -36 \][/tex]
Hence, [tex]\( f\left(\frac{1}{3}\right) = -36 \)[/tex].
2. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 3 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 3 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{10}{3} \)[/tex].
- Plug [tex]\( x = \frac{10}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{10}{3}\right) = 9\left(\frac{10}{3}\right)^2 - 54\left(\frac{10}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{10}{3}\right) = 9 \cdot \left(\frac{100}{9}\right) - 54 \cdot \left(\frac{10}{3}\right) - 19 = 100 - 180 - 19 = -99 \][/tex]
Hence, [tex]\( f\left(\frac{10}{3}\right) = -99 \)[/tex].
3. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 6 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 6 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{19}{3} \)[/tex].
- Plug [tex]\( x = \frac{19}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{19}{3}\right) = 9\left(\frac{19}{3}\right)^2 - 54\left(\frac{19}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{19}{3}\right) = 9 \cdot \left(\frac{361}{9}\right) - 54 \cdot \left(\frac{19}{3}\right) - 19 = 361 - 342 - 19 \approx 0 \][/tex]
Hence, [tex]\( f\left(\frac{19}{3}\right) \approx 0 \)[/tex].
4. Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 9 \frac{1}{3} \)[/tex]:
- Convert [tex]\( x = 9 \frac{1}{3} \)[/tex] to an improper fraction [tex]\( x = \frac{28}{3} \)[/tex].
- Plug [tex]\( x = \frac{28}{3} \)[/tex] into the function:
[tex]\[ f\left(\frac{28}{3}\right) = 9\left(\frac{28}{3}\right)^2 - 54\left(\frac{28}{3}\right) - 19 \][/tex]
- Simplify:
[tex]\[ f\left(\frac{28}{3}\right) = 9 \cdot \left(\frac{784}{9}\right) - 54 \cdot \left(\frac{28}{3}\right) - 19 = 784 - 504 - 19 = 261 \][/tex]
Hence, [tex]\( f\left(\frac{28}{3}\right) = 261 \)[/tex].
By evaluating the function [tex]\( f(x) \)[/tex] at these points, we observe that:
- [tex]\( f\left(\frac{1}{3}\right) = -36 \)[/tex]
- [tex]\( f\left(\frac{10}{3}\right) = -99 \)[/tex]
- [tex]\( f\left(\frac{19}{3}\right) \approx 0 \)[/tex]
- [tex]\( f\left(\frac{28}{3}\right) = 261 \)[/tex]
Therefore, the value [tex]\( x = 6 \frac{1}{3} \)[/tex] (or [tex]\( x = \frac{19}{3} \)[/tex]) is a zero of the quadratic function [tex]\( f(x) = 9x^2 - 54x - 19 \)[/tex].
