Question

please can someone help me ?

Answer 1

where is ur Question ?!!

Answer 2

The graph of which equation is shown below?

Answer 3

parabola

Answer 4

vertex is at \((2,-3)\) so it must look like
\[y=a(x-2)^2-3\] all we need is \(a\)

Answer 5

how do i get a ?

Answer 6

the curve passes through 4,1
i.e put y=1 when x=4 to get a

Answer 7

we know from the graph that if \(x=0\) you get \(y=1\) because \((0,1)\) is on the graph
so replace \(x\) by 0, set the result equal to 1 and solve for \(a\)

Answer 8

i think we can take a point frm X & another frm Y
then we hav 1 variable & we can get a

Answer 9

\[1=a(0-2)^2-3\]
\[1=4a-3\]
\[4=4a\]
\[a=1\]

Answer 10

and so your answer is
\[y=(x-2)^2-3\] or if you like
\[y=x^2-4x+1\]

Answer 11

okay thankyou so much!

Answer 12

yw