Question

for the parabola x^2+2x+28y-111=0. find the equation of the axis and what is the maximum value ?

Answer 1

yeah thanks i got it :)!

Answer 2

seperate x and y terms
x^2+2x=111-28y
complete square for x terms
x^2+2x+1-1=111-28y
(x+1)^2=112-28y
writing in standard form
(x+1)^2=-28(y-4)

compare with standard equation
(x-h)^h=4p(y-k)
for this standard parabola x=h is it axis

so
x=-1 is the parabola axis

and for maximium vale one method is to take derivative ( i assume you dont know it yet ) . other way is since this parabola is concave down so the max value is the y value of the vortex which is 4 so y=4 is the maximium hieght.

Answer 3

welcome .:)