Question 1173305
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Hi  
Note: the vertex form of a Parabola opening up(a>0) or down(a<0), 
{{{y=a(x-h)^2 +k}}} 
where(h,k) is the vertex  and  x = h  is the Line of Symmetry.
y=2x^2+12x+65  0r Completing the Square
y= 2(x^2+6x +9) -18 + 65  
y= 2(x+3)^2) + 47
V(-3,47)  and x = -3 is the Line of Symmetry
Minimum Value is at  P(-3,47) 
{{{drawing(300,300,    -10,10,-100,100,  blue(line(-3,100, -3,-100))  ,  
 grid(1),
circle(-3, 47,0.4),
graph( 300, 300, -10,10,-100,100,0, y=2x^2+12x+65))}}}
y=-7x^2+14x+3 0r Completing the Square
y = -7(x^2 - 2x +1)+7 +3
y =(x-1)^2-4  C(1, 10)  and x = 1 is the Line of Symmetry
Maximum Value is at  P(1, 10)
Wish You the Best in your Studies.
{{{drawing(300,300,    -20,20,-20,20,  blue(line(1,20, 1,-20))  ,  
 grid(1),
circle(1, 10,0.4),
graph( 300, 300, -20,20,-20,20, 0, -7x^2+14x+3  ))}}}