How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=3x^2+6x-1#?

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Answer 1 · 2017-03-08T06:24:46

Y-intercept #(0, 1)#

Vertex #(-1, -2)#

Given -

#y=3x^2+6x-1#

To calculate y-intercept put #x=0#

#y=3(0)^2+6(0)-1#

#y=1#

Vertex

#x=(-b)/(2a)=(-6)/(2xx3)=(-6)/(6)=-1#

At #x=-1/2#

#y=3(-1)^2+6(-1)+1=3-6+1=-2#

Vertex #(-1, -2)#
.
Graph

Answer 2 · 2017-03-08T06:25:09

Y-intercept #(0, 1)#

Vertex #(-1, -2)#

Given -

#y=3x^2+6x-1#

To calculate y-intercept put #x=0#

#y=3(0)^2+6(0)-1#

#y=1#

Vertex

#x=(-b)/(2a)=(-6)/(2xx3)=(-6)/(6)=-1#

At #x=-1/2#

#y=3(-1)^2+6(-1)+1=3-6+1=-2#

Vertex #(-1, -2)#

Table
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.
Graph