Hi,

f(x)=2x^2+4x+5  y=intercept Pt(0,5)

the vertex form of a parabola,  where(h,k) is the vertex

f(x)=2[(x^2 +2x} +5

f(x)=2[(x+ 1)^2 -1} +5

f(x)=2(x+ 1) -2 + 5

f(x)=2(x+ 1)+3

vertex Pt(-1,3), parabola opens upward (a=2 and is > than zero)





 


 
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation  (in our case ) has the following solutons:

  

  

  

  For these solutions to exist, the discriminant  should not be a negative number.

  

  First, we need to compute the discriminant : .

  

  The discriminant -24 is less than zero. That means that there are no solutions among real numbers.


  If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


  

      In the field of imaginary numbers, the square root of -24 is + or - . 

  

      The solution is 

  

  Here's your graph:





 

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