Could someone please help me?
@Atrineas
Uhh well the vertex has a fraction so I don't know how to do that but I have the Y inter answer and X inter Answer
@TheSmartOne can you help get the vertex please
mm I don't know either but I'll brb I've got some things to do it may to a long time or not but I'll be back as soon as I can
ok
@rebeccaxhawaii can you help her get the vertex plz for the equation -5x^2-2x-4
sure
K thanks
@rebeccaxhawaii it has to be entered in as a fraction which is why I'm having trouble
That what @Atrineas said and I can't see them website @rebeccaxhawaii my laptop doesn't allowed me too
It's basically takes you to desmos
Oh
A quadratic function is a function of the form f(x)=ax^2+bx+c, where a, b, and c are real numbers and a = 0 for the given function a=-5, b=-2 and c=-4
Okay so it looks like we're trying to get factions for this one
The vertex occurs at x= \[-\frac{ b }{ 2a }\] To find the x-coordinate of the vertex of the given function, substitube the values of a and b into the expression and simplify. Let a=-5 and b= -2 \[\frac{ -2 }{ 2(-5) }=-\frac{ 2 }{ 10 }\]
Idk maybe. . .
Okay say that \[-\frac{ 2 }{ 10}\] is the x-coordinate, we would evaluate to find the y-coordinate r \[r(-\frac{ 2 }{ 10 })=-5(-\frac{ 2 }{ 10 })^2-2(-\frac{ 2 }{ 10 })-4=?\]
yes
okay but what would the second part be?
This is for vertex right?
yea
ok so x=-2/10 and y=? The answer to the last problem you just wrote with all the fractions which Equals Whatever it equals
The one I got wrong?
so um\[(-\frac{ 2 }{ 10 }, 3)?\]
Or would it be \[(-\frac{ 2 }{ 10 }, -2)\]
Ehto I think y=-4
It says its wrong
I think I figured it out not 100% sure tho
|dw:1462683671130:dw| sorry if you can't read it. It was just something I was thinking still don't know if it 100% right
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