Question

a(find the vertex of the parabolab)is the vertex a minimum or a maximum

Answer 1

The equation of a parabola in vertex form is given by:

$y=a(x-h)^2+k$

where h and k are the coordinates of the center of the vertex

The given equation

$y=2x^2+4x-5$

can be expressed in vertex form by following the steps:

Step1: factor out 2 to get a

$y=2(x^2+2)-5$

Step2: find the square of half of 2

$\Rightarrow1^2$

Step 3: Re-write the equation

$\begin{gathered} y=2(x^2+2+1^2)-1^2-5 \\ y=2(x+1)^2_{}-1^2-5 \end{gathered}$$y=2(x+1)^2-6$

Thus the equation of the vertex is

$\begin{gathered} y=2(x+1)^2-6 \\ \end{gathered}$

If we compare this with the equation of a parabola in vertex form

$\begin{gathered} a=2 \\ h=-1 \\ k=-6 \end{gathered}$

From the values given

Part A

The vertex of the parabola is

$\begin{gathered} \Rightarrow\text{ (-1,-6)} \\ \Rightarrow h=-1,\text{ k=-6} \end{gathered}$

Part B

From the equation

a=2

Since the value of a is a positive, The vertex is a minimum