Question

need help ill give metals

Answer 1

Which system of equations is represented by the graph?

Answer 2

@DanJS
@SolomonZelman

Answer 3

well first do you know how to find the equation for the line?

Answer 4

y = x2 – 6x – 7x – y = 1
y = x2 – 6x + 7x + y = –1
y = x2 + 6x – 7x – y = 1
y = x2 + 6x + 7x + y = 1

Answer 5

no

Answer 6

Those equations don't make any sense...

Answer 7

anyways you can find the slope of the line by using the slope formula

Answer 8

\[m=\frac{y_2-y_1}{x_2-x_1}\]

Answer 9

then about you find the slope
use the slope-point form for line
\[y-y_1=m(x-x_1) \text{ this will be the equation of the line } \\ \text{ where you have already found m and identify one point } (x_1,y_1) \text{ on the line }\]

Answer 10

@freckles for the first on i got 3/3=1

Answer 11

5-(-2)=5+2 isn't it
but yeah the slope is1
because you have 7/7=1

Answer 12

so you have m=1
and you can use either point for the (x1,y1) thing

Answer 13

so what is your equation for the line?

Answer 14

would the answer be y=x+10-x-4
Y=x

Answer 15

did you use the \[y-y_1=m(x-x_1) \\ \text{ where } m=1 \\ y-y_1=1(x-x_1) \\ y-y_1 =(x-x_1) \\ y-y_1=x-x_1\]
where your job now is to just enter in either (5,5) or (-2,-2) to get the equation for the line

Answer 16

\[(x_1,y_1)=(5,5) \text{ replace } x_1 \text{ with 5 and } y_1 \text{ with 5 }\]

Answer 17

if i do that wouldnt that come out to zero

Answer 18

nope

Answer 19

unless you are talking about the y-intercept?

Answer 20

yes because 5-5=0*1=0

Answer 21

how did you get that

Answer 22

all I'm asking you to do is replace the x_1 and the y_1

Answer 23

it looks you are replacing everything with 5

Answer 24

including the things that vary

Answer 25

so it would be y-5=x-5?

Answer 26

perfecto

Answer 27

add 5 to both sides and you have
y=x-5+5
y=x

Answer 28

now the other equation is going to be a little more difficult so I will point some things out
I notice the other graph has a vertical asymptote at x=4
so I know for the other equation I could probably have something that looks like:
\[y=\frac{a}{x-4}+h\]
We can find a and h by using the points given

Answer 29

would a be -10?

Answer 30

The thing is when I find a and h
the equation I get gives a different horizontal asymptote than the graph given

Answer 31

So I guess we need to go back to try to understand what you were saying by the equations given

Answer 32

For example,
I can't read or understand this:
y = x2 – 6x – 7x – y = 1
y = x2 – 6x + 7x + y = –1
y = x2 + 6x – 7x – y = 1
y = x2 + 6x + 7x + y = 1

Answer 33

the choices are

y=-x-10/x-4
y=x

Y=x+10/x-4
y=-x

y=-x+10/x+4
y=x

y=-x+10/x-4
y=-x

@freckles

Answer 34

we know it is either A or C

Answer 35

I can't totally tell what your equations say.
Like if it is y=(-x-10)/(x-4) or if it is y=-x-10/(x-4)

Answer 36

i wanna say its A

Answer 37

but you can plug in the points

Answer 38

like see what happens when you plug in (5,5)

Answer 39

are both sides the same?

Answer 40

idk how to do that

Answer 41

(x=5,y=5)
so replace x with 5 and y with 5

Answer 42

and compare both sides of the equation when done

Answer 43

thats why i said A but then it might be C so idk

Answer 44

also A should be right because it makes sense for the bottom to be x-4
but I'm concerned if it is A
I'm trying to vertify the points (5,5) and (-2,-2) are actually on the equation
But I'm also having trouble reading the equation.
\[y=\frac{-x-10}{x-4} \\ 5=\frac{-5-10}{5-4} \\ 5=\frac{-15}{1} \\ \text{ \not true so I don't think I'm reading the equation correctly } \\ \text{ maybe \it is } y=-x-\frac{10}{x-4} \\ \text{ plug \in } (5,5) \\ 5=-5-\frac{10}{5-4} \\ 5=-5-\frac{10}{1} \\ 5=-5-10 \\ 5=-15 \\ \text{ also \not true }\]

Answer 45

is it a or c cuz that just confused me

Answer 46

Is C a miss-typed because A doesn't work anyway it can be interpreted

Answer 47

thanx