Determine

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|dw:1443540278601:dw|
thats a parabola sorry it was hard to draw

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PLEASE HELPPPP @Nnesha
\[\huge\rm y=a(x-h)^2+k\] vertex form where (h,k) is the vertex point so substitute (h,k) for (-2,-4)
I did that and i got y=a(x+2)^2-4 @Nnesha
right the leading coefficient makes the graph wider or skinnier or turn up and down (depends on the sign ) if a is negative then it opens down
but i am confused on which one my answer will be
why ?? you already got the equation
I'm thinking A?
leading coefficient is a
why A ?
because where does the "a" i the equation go? and then how do i tell if it is negative or not
because i got y=a(x+2)^2-4 and the closest one to that is A
\[\huge\rm y=a(x-h)^2+k\] vertex form a outside the parentheses
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha right the leading coefficient makes the graph wider or skinnier or turn up and down (depends on the sign ) if a is negative then it opens down \(\color{blue}{\text{End of Quote}}\) read this
but its going up?
but it opens down...
\(\color{blue}{\text{Originally Posted by}}\) @iwanttogotostanford because i got y=a(x+2)^2-4 and the closest one to that is A \(\color{blue}{\text{End of Quote}}\) right you got this now look at first option both are the same or not ?
no, because it i missing the A
well look at option A there is +4 but in ur equation it's -4
ah ok so it would be option D @Nnesha
yes right and bec the graph opens down that's how there is negative sign
okay, thank you
|dw:1443541832026:dw|
np
can i ask you another graph question with a new problem please?
i'll try not sure i'm not good at graphing stuff

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