\begin{tabular}{l} N. 11 Write a quadratic function from ita vertax and anothar point Univ \\ A parabola opening up or down has vertex \( (-1,-4) \) and passes through \( \left(-18, \frac{209}{20}\right) \). Write its \\ equation in vertex form. \\ Simplify any fractions. \\ \hline\end{tabular}

To write the equation of a quadratic function in vertex form, we use the formula: \[ f(x) = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola. In this case, the vertex is given as \((-1, -4)\), so we have: \[ f(x) = a(x + 1)^2 - 4 \] Next, we need to find the value of \(a\). We can do this by using the other point that the parabola passes through, which is \(\left(-18, \frac{209}{20}\right)\). We will substitute \(x = -18\) and \(f(x) = \frac{209}{20}\) into the equation to solve for \(a\). Substituting these values into the equation: \[ \frac{209}{20} = a(-18 + 1)^2 - 4 \] This simplifies to: \[ \frac{209}{20} = a(-17)^2 - 4 \] Calculating \((-17)^2\): \[ (-17)^2 = 289 \] So we have: \[ \frac{209}{20} = 289a - 4 \] Next, we add 4 to both sides. To do this, we convert 4 to a fraction with a denominator of 20: \[ 4 = \frac{80}{20} \] Thus, we have: \[ \frac{209}{20} + \frac{80}{20} = 289a \] This simplifies to: \[ \frac{289}{20} = 289a \] Now, we can solve for \(a\) by dividing both sides by 289: \[ a = \frac{289}{20} \div 289 = \frac{1}{20} \] Now that we have \(a\), we can write the equation of the quadratic function in vertex form: \[ f(x) = \frac{1}{20}(x + 1)^2 - 4 \] Thus, the final equation of the quadratic function is: \[ \boxed{f(x) = \frac{1}{20}(x + 1)^2 - 4} \]