Answer
Choice F
Work Step by Step
To find the vertex of the given equation, $
y=2x^2+4x-3
,$ convert the equation in the form $y=a(x-h)^2+k$.
Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
y&=(2x^2+4x)-3
\\
y&=2(x^2+2x)-3
.\end{align*}
Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to
\begin{align*}
y&=2\left(x^2+2x+\left(\dfrac{2}{2}\right)^2\right)+\left[-3-2\left(\dfrac{2}{2}\right)^2\right]
\\\\
y&=2\left(x^2+2x+1\right)+\left[-3-2\right]
\\
y&=2\left(x+1\right)^2-5
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
2\left(\dfrac{2}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of the equation $y=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic equation above is $
\left(-1,-5\right)
$. Thus, the vertex is located on the third quadrant.
In the equation $
y=2\left(x+1\right)^2-5
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=2$ (greater than zero), then the parabola opens upwards.
With the vertex located at Quadrant $3$ and the parabola opens up, then the corresponding graph of the given equation is Choice F.
