dit View History Bookmarks Window Help D G h Dashboard | Carmel High School Unit 4 Test Started: Jul 6 at 1:19pm Quiz Instructions Question 3 indianaonline.instructure.com indianaonline.instructure.com ל 3 Quiz: Unit 4 Test Questions Identify the quadratic function $y=x^{2}+10 x+24$ in vertex form, identify the vertex, and choose whether the vertex is a maximum or minimum. vertex $(-5,-1)$ is a minimum vertex $(-5,-1)$ is a maximum vertex $(5,-1)$ is a maximum $y=(x+5)^{2}+1$ $y=(x-5)^{2}+1$ $y=(x-5)^{2}-1$ $y=(x+5)^{2}-1$ vertex $(5,-1)$ is a minimum Time Elapsed: Attempt due: Jul 2 a 50 Minutes, 33 i $\ddot{3}$ $: 8$ || ${ }_{5}^{*}|| \hat{6}$ \begin{tabular}{l||l||l} 8 & i \\ \hline \end{tabular} \% $E\left[\begin{array}{l|l|l|l} & T & Y & U\end{array}\right.$ $\boldsymbol{s}$ D G H
Solution
Step 1 ：The vertex form of a quadratic function is given by \(y=a(x-h)^{2}+k\), where \((h,k)\) is the vertex of the parabola.
Step 2 ：The vertex of a parabola \(y=ax^{2}+bx+c\) is given by \(-\frac{b}{2a}, f(-\frac{b}{2a})\).
Step 3 ：Given the quadratic function \(y=x^{2}+10x+24\), we can find the vertex by substituting the coefficients into the formula. Here, \(a=1\) and \(b=10\).
Step 4 ：Calculating the vertex, we get \(h=-\frac{b}{2a}=-5\) and \(k=f(-\frac{b}{2a})=-1\).
Step 5 ：Since the coefficient of \(x^{2}\) is positive, the parabola opens upwards, and the vertex is a minimum.
Step 6 ：The vertex form of the function is \(y=(x+5)^{2}-1\), and the vertex is \((-5,-1)\).
Step 7 ：\(\boxed{\text{Final Answer: The vertex form of the function is } y=(x+5)^{2}-1, \text{ the vertex is } (-5,-1), \text{ and the vertex is a minimum.}}\)
