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what do you think?
do i solve the problem?
All you have to do is find out if it's true or false
There are a couple different ways you could approach this. The equation is already provided, you just need to know if it's the correct equation for the graph. You could pick points on the graph and plug them into the equation to see if they fit, or just think back to the vertex form equation...
i know, but how do i know? like should i solve the problem for x to see if its right? online schooling gets confusing
woo, getting more confussed, to many people replying, my computer being slow to catch up lol
no it's true plug in (3,1) \(f(x) = 3(x - 3)^2 + 1\\1=3(3-3)^2+1\\3-3=0\\0^2=0\\3*0=0\\1=1\) another point (2,4) \(4= 3(2 - 3)2 + 1\\4=3*-1^2+1\\-1^2=1\\3*1=3\\3+1=4\\4=4\)
@sohailiftikhar is wrong
i see what you mean @Mehek14 .
@Penguin7 , I agree with @Mehek14 !
how am I wrong when the solutions are correct?
if you want more proof, why don't you graph it using desmos?
desmos confuses me..just saying lol
because that is one of the points in the parabola
so it has to be a solution in the equation if the equation is true
the output of x=1 is not shown in the graph that @Penguin7 attached
Using the vertex form equation, -h is how far the graph moves to the right and +k is how far up it moves. Your graph moved 3 to the right and 1 up so that gets you to y=a(x-3)^2+1
to find a, you can plug in values...(2,4) and (4,4) are clearly visible on the graph.
you have (3,1), (2,4), and (4,4) in both graphs
so, can we all agree that @Mehek14 is right?
@Penguin7 , yes!!! @Mehek14 is correct. If it helps you decide at all, I am a math teacher. :)
okay. well, i hope both you and @Mehek14 are okay will me taging u both..bc i know ill need help.