Vertex of a Parabola

$$ y=(x+2)^2+1 $$



$$ y=-(x-2)^2-1$$



$$ y=-(x+2)^2-1 $$



$$ y=-(x-2)^2+1 $$



$$ y=-(x+2)^2+1 $$



$$ y=(x-2)^2-1 $$



$$ y=(x+2)^2-1 $$



$$ y=(x-2)^2+1 $$

Example 1

Find the vertex of $y=x^2 + 4x +2$.

\( \begin{align} \displaystyle
y &= x^2 + 4x +2 \\
&= x^2 + 4x + 4 – 2\\
&= (x+2)^2-2 \\
\therefore &(-2,-2) \\
\end{align} \)

Example 2

Find the vertex of $y=x^2 – 6x +1 $.

\( \begin{align} \displaystyle
y &= x^2 – 6x +1 \\
&= x^2 – 6x + 9 +8 \\
&= (x-3)^2 +8 \\
\therefore &(3,8) \\
\end{align} \)

Example 3

Find the vertex of $y= -x^2 + 4x +1 $.

\( \begin{align} \displaystyle
y &= -(x^2 -4x) + 1 \\
&= -(x^2 -4x + 4) + 4 + 1 \\
&= (x-2)^2+5 \\
\therefore &(2,5) \\
\end{align} \)





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