I have a pre-calculus inquiry
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I have a pre-calculus inquiry

[From: ] [author: ] [Date: 12-03-02] [Hit: ]
-Your calculations seem to be impeckable but the french should be : Silvia go splatt? Mercy!But that may be the cajun translation and not real Canadian French and certainly not Froggie type.-well if the vertex is (-2,......
Alright so...

y= -(x+2)^2.

Could someone tell me whether or not I got these wrong?
For the vertex: (-2,0)
Axis of Symmetry or line of symmetry: x=-2
Maximum: y=0
y-intercept= y= (0,-4)

and for the x-intercept, I know I got this wrong, so I'll need someone to help me out. S'il vous plait, merci.

y= -(x+2)^2
0= -(x+2)^2
0= -(x^2+4x+4)
0= -x^2 -4x -4

and I have no clue.

-
the x-intercept in this case is the same as the vertex (-2,0)

If you have a polynomial in the form ax^2 + bx+c = 0, you factor it and the "solutions" are your x-intercepts.

In this case, you already have it factored -(x+2)^2, so you know that the x-intercept is -2 just by looking at it.

-
Your calculations seem to be impeckable but the french should be : Silvia go splatt? Mercy!
But that may be the cajun translation and not real Canadian French and certainly not Froggie type.

-
well if the vertex is (-2,0) then there will be no x intercepts
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