How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=3x^2+10x#?

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Answer 1 · 2016-11-13T12:43:26

#y#-intercept is #0#, axis of symmetry is #x+5/3=0# and vertex is #(-5/3,-25/3)#

#y#-intercept is given by #y=f(0)=3xx0^2+10xx0=0#,

hence, #y#-intercept is #0#.

Now #y=f(x)=3x^2+10x#

= #3(x^2+10/3x+(5/3)^2)-3xx(5/3)^2#

= #3(x+5/3)^2-25/3#

Hence, axis of symmetry is #x+5/3=0#

and at #x=-5/3#, #y=f(x)=-25/3#

Hence, vertex is #(-5/3,-25/3)#
graph{3x^2+10x [-21.08, 18.92, -11.28, 8.72]}