412brooke
contestada

Where is the vertex of the parabola? y = x2 + 6x + 9
a) on the y-axis
b) above the x-axis
3) below the x-axis
4) on the x-axis

Respuesta :

[tex] \bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+6}x\stackrel{\stackrel{c}{\downarrow }}{+9}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{6}{2(1)}~~,~~9-\cfrac{6^2}{4(1)} \right)\implies (-3~~,~~9-9)\implies (-3,0) [/tex]


now that point is right on the x-axis, right between the II and III Quadrants.

gmany

[tex]f(x)=ax^2+bx+c\\\\\text{The formula of a vertex:}\\(h,\ k)\ \text{where}\ h=\dfrac{-b}{2a}\ \text{and}\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[/tex]

[tex]\text{We have}\\\\f(x)=x^2+6x+9\to a=1,\ b=6,\ c=9[/tex]

[tex]\text{Substitute:}\\\\h=\dfrac{-6}{2\cdot1}=\dfrac{-6}{2}=-3\\\\k=f(-3)=(-3)^2+6(-3)+9=9-18+9=0[/tex]

[tex]\text{The vertex}\ (-3,\ 0)\ \text{therefore your answer is}\\\\\boxed{4)\ on\ the\ x-axis}[/tex]


[tex]\text{Other method.}\\\\\text{The vertex formula of a quadratic function}\\\\f(x)=a(x-h)^2+k[/tex]

[tex](h,\ k)-vertex[/tex]

[tex]\text{We have}\ y=x^2+6x+9=\underbrace{x^2+2\cdot x\cdot3+3^2}_{(a+b)^2=a^2+2ab+b^2}=(x+3)^2=(x-(-3))^2+0\\\\\text{therefore}\ h=-3,\ k=0[/tex]

[tex]The\ vertex\ is\ in\ (-3,\ 0)-on\ the\ x-axis[/tex]

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