How do you find the vertex, axis of symmetry and intercepts of #f(x)=x^2+3x+7#?

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Answer 1 · 2018-01-16T06:03:01

Vertex #(-3/2, 19/4)#

Axis of symmetry #x=-3/2#

Given -

#y=x^2+3x+7#

Vertex
x coordinate of the vertex

#x=(-b)/2a=(-3)/(2 xx 1)=-3/2#
y coordinate of the vertex
At #x=-3/2#
#y=(-3/2)^2+3(-3/2)+7#
#y=9/4-9/2+7=(9-18+28)/4=19/4#

Vertex #(-3/2, 19/4)#

Axis of symmetry #x=-3/2#