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Dashboard | Carmel High School
Unit 4 Test
Started: Jul 6 at 1:19pm
Quiz Instructions
Question 3
indianaonline.instructure.com
indianaonline.instructure.com
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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}
\%
$E\left[\begin{array}{l|l|l|l} & T & Y & U\end{array}\right.$
$\boldsymbol{s}$
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H

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 ：\(\boxed{\text{Final Answer: The vertex form of the function is } y=(x+5)^{2}-1, \text{ the vertex is } (-5,-1), \text{ and the vertex is a minimum.}}\)