How do you graph #y=3x^2-6x-2#?

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Answer 1 · 2018-06-13T01:53:44

graph{3x^2-6x-2 [-10, 10, -5, 5]}

#x#-intercepts: #1-sqrt15/3" "# and #" "1+sqrt15/3#

Vertex: #(1, -5)#

Positive (U shaped)

There are 2 parts to graphing a parabola:

1) Finding the vertex

You can find the #x#-value for the vertex with this formula:

#x= -b/(2a)#

For this equation, #b= -6# and #a= 3#. Then plug this #x# back into the original equation, #y=3x^2-6x-2#, to find the #y# value.

2) Finding the #x#-intercepts by plugging values for #a#, #b#, and #c# into the quadratic formula

#x= (-b+-sqrt(b^2-4ac))/(2a)#

#x= (-(-6)+- sqrt((-6)^2-4(3)(-2)))/(2*3)#

And because the first term, #3x^2#, is positive, we know that the parabola is shaped like a right-side-up cup, or a U shape.