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@Nnesha Did I graph these 3 points, right?
hmm \[\huge\rm y=a(x-h)^2+k\] vertex form of parabola where (h,k) is the vertex point and find x and y intercepts
x-intercept is a point where line intersect x-axis when y=0 so substitute y for 0 to find x intercept y-intercept is a point where line intersect y-axis when x=0 so substitute x for y to find y-intercept
The vertex is (4,1)
what are the next steps?
alright and bec a is negative in the question so we would get max point (graph opens down)
so max would be at ((4,1) |dw:1443814926678:dw|
now find x intercept
hmm highest degree is 2 so you should get 2 xintercepts 4 isn't correct
hmm how did you get that ?
I graphed it on my calculator
hmm make sure you typed the correct equation
Alright, I got x=0 and y=-63
yeah that's y-intercept now substitute y for 0 to find x-intercept
let me know if u don't know how to solve for x
alright wait a sec
\[\huge\rm 0=-4(x-4)^2+1\] do you know the quadratic formula ?
(x-4)^2 is same as (x-4)(x-4) foil it
familiar with the foil method ?
hmmm better use calculator instead google i didn't get 16
make sure you are looking at the right question
i have no idea how u got that please show some work we need x intercepts when y =0
-sighs- I typed it in my calculator. (x-4)^2 and graphed, it showed x=0 and y=16
hmm well ur equations is -4(x-4)^2 +1 that's what u should type into the calculator
lots of step lol
will if you want to solve by hand to find x-itnercepts then yes there are lots of steps if you want to use calculator then there are none
Alright, I typed it in. Graphed. X=0
lol i was about to post ur question on others thread :D
I'm sorry, I am not that good at math, I'm barely on an intermediate level.
or here is another way to graph x-coordinate of vertex is 4 right so draw a table |dw:1443819080897:dw| now substitute those points to find y coordinate
no i have so man tabs open so that's why
do you understand that ?
don't need to find x-intercepts just substitute x for the points in the table that's it
vertex is max point min = minimum max = maximum |dw:1443820566646:dw|
|dw:1443820718145:dw| if the vertex is at (4,1) that would be the maxx point
|dw:1443820739973:dw| this point should be the max that's why to find other two points we should go up 2 and down 4 from the x-coordinate of the vertex
|dw:1443820778704:dw| use this table substitute x for its value from the table then solve for y