Let \( f(x)=\frac{2 x^{2}-9 x+9}{3 x^{2}-14 x+8} \) This function has: 1) Ay intercept at the point \( \left(0, \frac{9}{8}\right) \) 2) \( x \) intercepts at the point(s) \( \left(\frac{3}{2}, 0\right),(3,0) \) 3) Vertical asymptotes at \( x= \)

To find the vertical asymptotes of the function \( f(x)=\frac{2 x^{2}-9 x+9}{3 x^{2}-14 x+8} \), we need to identify where the denominator equals zero. Setting the denominator \( 3 x^{2}-14 x+8 = 0 \) and solving for \( x \) will yield the points of vertical asymptotes. Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3, b = -14, c = 8 \), we find the roots: \[ x = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} \] \[ = \frac{14 \pm \sqrt{196 - 96}}{6} = \frac{14 \pm \sqrt{100}}{6} = \frac{14 \pm 10}{6} \] Thus, the roots are: \[ x = \frac{24}{6} = 4 \quad \text{and} \quad x = \frac{4}{6} = \frac{2}{3} \] This means the vertical asymptotes occur at \( x = 4 \) and \( x = \frac{2}{3} \). So, summarizing: - The vertical asymptotes are at \( x = 4 \) and \( x = \frac{2}{3} \). Now, let’s add more flavor to this mathematical journey! Have you ever pondered how ancient civilizations approached ratios and fractions? The ancient Egyptians had their unique fraction system, using only unit fractions (like \( \frac{1}{2} \), \( \frac{1}{3} \)) to represent values. They would reject using fractions like \( \frac{3}{7} \) directly! The reasoning? Simplicity and clarity, reminiscent of how we simplify algebraic functions today! And after diving into the world of mathematical functions, why not explore some fun applications? Rational functions like \( f(x) \) pop up everywhere from physics (think of ratios of forces) to economics (modeling cost-revenue relationships). Understanding these functions not only hones your algebra skills but also sets you on the path to conquering real-world scenarios – where math meets money and mechanics!