Question

Find the direct variation equation of the graph through the points (0, 0) and (1, -2). Write in y=kx form.
y = 2x
y = -2x
y = 1/2 x
y = -1/2 x

Answer 1

guess i didn't help the first time.
do you know what exactly you are looking for in this problem?

Answer 2

you are looking for a number, that they call "\(k\)" in this problem
your choices from the possible answers are either \(k=2,k=-2,k=\frac{1}{2}\) or \(k=-\frac{1}{2}\) and your job it to find the correct one

Answer 3

i just dont know how to change the form

Answer 4

the point \((0,0)\) is not much help to solve this, but the point \((1,-2)\) is because this tells you that if \(x=1\) then \(y=-2\)
is that clear or not?

Answer 5

in other words, you can find \(k\) by looking at the equation
\[y=kx\] replace \(x\) by \(1\) and \(y\) by \(-2\)

Answer 6

the first numkber or x is 1 and the second number or y is -2?

Answer 7

so -2=k1?

Answer 8

yes in that particular example
\(x\) and \(y\) are variables, but you know that IF \(x=1\) THEN \(y=-2\)

Answer 9

yes

Answer 10

and since \(k\times 1=k\) you know that \(-2=k\) which is what you were looking for

Answer 11

so now go back the the original equation
\[y=kx\] and where you see a "\(k\)" replace it by \(-2\)

Answer 12

yes

Answer 13

it im a little confused how do you end uo with k= instead of y=

Answer 14

you would write
\[1=k\times 4\] and solve for \(k\) by dividing both sides by \(4\) to get
\[\frac{1}{4}=k\]
then go back to
\[y=kx\] and replace \(k\) by \(\frac{1}{4}\)

Answer 15

yes i see that you are confused by that
\(k\) is some fixed number that you are trying to find
\(x\) and \(y\) are variables

Answer 16

thanks

Answer 17

so for example, once we know the equation is
\[y=-2x\] for the first example, \(x\) and \(y\) can vary, they are not fixed numbers
for example if \(x=3\) then \(y=-2\times 3=-6\) and if
\[x=5\] then
\[y=-2\times 5=-10\]

Answer 18

the points would look like \((3,-6)\) and \((5,-10)\)