I need help with this question (graphing quadratic functions);

y=x^2 + 4x + 8

Question

I need help with this question (graphing quadratic functions);

y=x^2 + 4x + 8

anonymous · Answer

Pls see http://www.youtube.com/watch?v=mDwN1SqnMRU

anonymous · Answer

First find the  vertex of this parabola  by using the formula 
( -b/2a , f( -b/ 2a ) )    this is the first point of this parabola  
then plug any value of x  and then find the corresponding value of  y  
and plot this points 
draw the smooth curve passing through that point this is the graph of this parabola
I hope this will help you ...

anonymous · Answer

since the coefficient of x^2 is +ve it will open up side parabola (U-Shape) put x=0 it will give where t intersect with y axis  lowest point of curve will be at x=(-b/a)