dealing with porabolas and quadratic equations...
i need help with this problem ,
y= -x^2 -3x =18
not sure if im doing it right, but when i tried solving for y, & to solve for y you would use -b/2a , i got 3/2.. am i right?

Question

dealing with porabolas and quadratic equations...
 i need help with this problem , 
  y= -x^2 -3x =18 
 not sure if im doing it right, but when i tried solving for y,  & to solve for y you would use -b/2a , i got 3/2.. am i right?

baldymcgee6 · Answer

the equation already has y isolated... -b/2a is to find the vertex

anonymous · Answer

oops the equation is y=-x^2-3x+18

anonymous · Answer

tht is what im trying to find, the vertex to plot the porabola

baldymcgee6 · Answer

okay, then you are on the right track with that formula, let me check...

baldymcgee6 · Answer

You should find the vertex to be at (-3/2, 20.25)

baldymcgee6 · Answer

Once you use the vertex formula to find the x value of the vertex, plug it back in to fund the y value

baldymcgee6 · Answer

-b/2a = -2/3

 y=-(-2/3)^2-3(-2/3)+18
y = 20.25

anonymous · Answer

so it should be a fraction. and yep thats what i got  thanks.!

baldymcgee6 · Answer

you're welcome