Question

Please help. BRAINLIEST TO BEST ANSWER

Answer 1

Answer:

b

Step-by-step explanation:

we have the point

$(2,-4)$

so when

x=2

y=4

so we can try on all equations

$y_1=2|x-2|-4\\\\y_1=2|2-2|-4\\\\y_1=2|0|-4\\\\y_1=0-4\\\\y_1=-4$

so equation 1 could be

$y_2=\frac{1}{2} |x-2|-4\\\\y_2=\frac{1}{2}|2-2|-4\\\\y_2=\frac{1}{2}|0|-4\\\\y_2=0-4\\\\y_2=-4$

so equation 2 could be

$y_3=2|x+2|-4\\\\y_3=2|2+2|-4\\\\y_3=2|4|-4\\\\y_3=8-4\\\\y_3=4$

so equation 3 is not

$y_4=\frac{1}{2}|x+2|-4\\\\y_4=\frac{1}{2}|2+2|-4\\\\y_4=\frac{1}{2}|4|-4\\\\y_4=\frac{1}{2}(4)-4\\\\y_4=2-4\\\\y_4=-2$

so equation 4 is not

So another point that we have is

$(0,-3)$

so we plug in

$y_1=2|x-2|-4\\\\y_1=2|0-2|-4\\\\y_1=2|-2|-4\\\\y_1=4-4\\\\y_1=0$

so this isn't

$y_2=\frac{1}{2}|x-2|-4\\\\y_2=\frac{1}{2}|0-2|-4\\\\y_2=\frac{1}{2}|-2|-4\\\\y_2=\frac{1}{2}(2)-4\\\\y_2=1-4\\\\y_2=-3$

so the answer is b

Answer 2

It’s B
I’m sure of it