QUestion 3
Given:
[tex]y=-x^2-14x-59[/tex]Let's find the vertex form of the equation.
To find the vertex form, take the vertex form of an equation:
[tex]y=a(x-h)^2+k[/tex]Apply the general form:
[tex]\begin{gathered} y=ax^2+bx+c \\ \\ y=-x^2-14x-59 \end{gathered}[/tex]Where:
a = -1
b = -14
c = -59
From the vertex equation, apply the formula below to find h:
[tex]h=\frac{b}{2a}[/tex]Substitute values into the formula:
[tex]\begin{gathered} h=\frac{-14}{2(-1)} \\ \\ h=\frac{-14}{-2} \\ \\ h=7 \end{gathered}[/tex]To find the value of k, apply the formula below:
[tex]k=c-\frac{b^2}{4a}[/tex]Substitute values into formula:
[tex]\begin{gathered} k=-59-\frac{-14^2}{4(-1)} \\ \\ k=-59-\frac{196}{-4} \\ \\ k=-59-(-49) \\ \\ k=-59+49 \\ \\ k=-10 \end{gathered}[/tex]From the vertex equation, substitute -1 for a, 7 for h and -10 for k.
We have:
[tex]\begin{gathered} y=-1(x-h)^2+k \\ \\ y=-1(x-7)^2+(-10) \\ \\ y=-(x-7)^2-10 \end{gathered}[/tex]Therefore, the vrtex form of the given equation is:
[tex]y=-(x-7)^2-10[/tex]ANSWER:
[tex]y=-1(x-7)^2-10[/tex]