Question

find the vertex,line of symetry, and minimum maximum value of the equation
f(x)=2x^2+2x+3

Answer 1

get first derivative

Answer 2

i have x = -1/2
y=5/2
line of symetry =5/2
and minimum =-1/2

Answer 3

get first derivative

Answer 4

and how did you go about findingthat

Answer 5

-b/2a=-2/4=-1/2 this is the vertex
f(-1/2)=2(-1/2)^2+3=5/2
line of symetry =5/2
and minimum =-1/2

Answer 6

hey awesome, good job
however, you need to switch your line of symmetry and minimum
max/min refer to the y-value
remember line of symmetry is that vertical line that cuts through the vertex

Answer 7

@ mathcruncher, this is not for a calculus class

Answer 8

you can also do it by completing the square
=2(x^2+x+3/2)
=2((x+0.5)^2-1/4+3/2)
=2((x+0.5)^2+5/4)
=2((x+0.5)^2)+5/2
line of symmetry will be equating (x+0.5)=0
therefore x=-0.5
x coor of vertex will be -0.5and y-coor will be subs x=-0.5 in the equation which is 5/2
it is a minimum bcoz the coeff of x^2 is positive, min at x=-0.5 adn min value will be -5/2