I have to find the vertex of: y= -3x^2+12x-8 and I have to plot it on a graph, providing an ordered pair as the vertex. I'm not understanding this at all. How should I begin?
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complete the sqaure
first factor out x^2 coeffiecnet
y = -3 ( x^2 -4x ) -8
note, we only factored out common factors on the x terms
then complete the sqaure in side the brackets
ok, it kinda makes sense so far..
to complete the square on x^2 -4x , we need to half the -4 ( to get -2 ) , the square it , and add and subtract inside the brackets
so y= -3 ( x^2 -4x + 4 - 4 ) - 8 // notice, its important you add and substract INSIDE THE BRACKET
we need to maintain equality
alright, with you on that
now you group the perfect sqaure inside the bracket
so y= -3 ( (x-2)^2 -4) -8
now you expand the brackets , being careful with the signs
so y= -3(x-2)^2 +12 -8
So I'm seeing here the only way the vertex can be solved is by doing the equation, then plotting it on the graph- how would I plot this #1, and #2, how would I come up with a solution set??
y= -3(x-2)^2 +4
vertex = (2,4)
you could do it other ways , you might the equation of axis of symmetry of a parabola is x=-b/2a
And how to plot? I have to do it on an online graph. Totally confused with it
and you know the vertex is on the axis of symmetry
I dont know there, I know how to do a general plot but have idea how the computer with mark you
"graph the equation by hand; find the vertex and round approximate coordinates to the second decimal place."
"Use the graphing tool to graph the parabola"
THIS is what is confusing me. Do I graph the solution set, I'm just not getting it. I understand for the most part the equation, but not plotting
One more equation- (x+8)^2+8(x+7)=64. I can solve by the method of my choice
Its not a negative sign in front of the x, I just need a solution set
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