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{{ printedBook.courseTrack.name }} {{ printedBook.name }} To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens *upward* or *downward*. To do so, we will first express it in vertex form and $k$ are either positive or negative numbers.
$y=2(x+1)_{2}−2⇔y=2(x−(-1))_{2}+(-2) $
Let's compare the general formula for the vertex form with our equation.
$General formula:y=Equation:y= a(x−-(h)_{2}+--k2(x−(-1))_{2}+(-2) $
We can see that $a=2,$ $h=-1,$ and $k=-2.$ Since the vertex of a quadratic function written in vertex form is the point $(h,k),$ the vertex of our function is $(-1,-2).$ Let's now determine the direction of the parabola.Recall that, if $a>0,$ the parabola opens *upwards.* Conversely, if $a<0,$ the parabola opens *downwards.*

In the given function, we have $a=2,$ which is *greater than* $0.$ Therefore, the parabola opens *upward*. As a result, the graph of the function is the graph given in choice **B**.