To determine an equivalent function to \( f(x) = 6x^2 - 23x + 21 \), we can try to factor the quadratic expression or complete the square.
First, let's factor \( f(x) \):
- We need to find two numbers that multiply to \( 6 \cdot 21 = 126 \) and add to \( -23 \).
- The numbers that work are \( -18 \) and \( -7 \), since \( -18 + -7 = -25 \) does not work but they are too large compared to \(-23\).
- Instead of focusing on direct factors, we can apply the quadratic formula to find the roots: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 6, b = -23, c = 21 \).
Substituting in the values: \[ x = \frac{-(-23) \pm \sqrt{(-23)^2 - 4 \cdot 6 \cdot 21}}{2 \cdot 6} = \frac{23 \pm \sqrt{529 - 504}}{12} = \frac{23 \pm \sqrt{25}}{12} = \frac{23 \pm 5}{12} \]
This gives us two solutions: \[ x = \frac{28}{12} = \frac{7}{3} \quad \text{and} \quad x = \frac{18}{12} = \frac{3}{2} \]
Thus, we can write \( f(x) \) in its factored form as follows: \[ f(x) = 6(x - \frac{7}{3})(x - \frac{3}{2}) \]
To express it simply, we can convert these back: \[ f(x) = 6 \left( x - \frac{7}{3} \right) \left( x - \frac{3}{2} \right) = 6 \left( \frac{3x - 7}{3} \right) \left( \frac{2x - 3}{2} \right) \]
So one equivalent function to \( f(x) = 6x^2 - 23x + 21 \) can also be given in factored form or the original form itself. The key is that the function remains the same regardless of the format you choose to express it in:
\[ f(x) = 6 \left( x - \frac{7}{3} \right) \left( x - \frac{3}{2} \right) \]
Or it can also be shown as: \[ f(x) = 6x^2 - 23x + 21 \]
These representations are all equivalent to each other.