Question

help, medal and fan
The following is the graph of f(x) = 3(x - 3)2 + 1.

Answer 1

what do you think?

Answer 2

do i solve the problem?

Answer 3

All you have to do is find out if it's true or false

Answer 4

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...

Answer 5

i know, but how do i know? like should i solve the problem for x to see if its right?
online schooling gets confusing

Answer 6

woo, getting more confussed, to many people replying, my computer being slow to catch up lol

Answer 7

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)
\(4= 3(2 - 3)2 + 1\\4=3*-1^2+1\\-1^2=1\\3*1=3\\3+1=4\\4=4\)

Answer 8

@sohailiftikhar is wrong

Answer 9

i see what you mean @Mehek14 .

Answer 10

@Penguin7 , I agree with @Mehek14 !

Answer 11

how am I wrong when the solutions are correct?

Answer 12

if you want more proof, why don't you graph it using desmos?

Answer 13

desmos confuses me..just saying lol

Answer 14

because that is one of the points in the parabola

Answer 15

so it has to be a solution in the equation if the equation is true

Answer 16

the output of x=1 is not shown in the graph that @Penguin7 attached

Answer 17

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

Answer 18

to find a, you can plug in values...(2,4) and (4,4) are clearly visible on the graph.

Answer 19

this matches the graph that was attached

Answer 20

you have (3,1), (2,4), and (4,4) in both graphs

Answer 21

so, can we all agree that @Mehek14 is right?

Answer 22

@Penguin7 , yes!!! @Mehek14 is correct. If it helps you decide at all, I am a math teacher. :)

Answer 23

okay. well, i hope both you and @Mehek14 are okay will me taging u both..bc i know
ill need help.