How do I reflect this over y=1?
2 answers:
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Hello once again!
When you see a question like this, you need to find the equation of the straight line.
The formular used is y = mx + c
Where
m = slope
c = constant
First find the slope, since it's a straight line, any 2 coordinates can be used.
Now we need to substitude in the slope, and one of the coordinate you used to find the slope, to the formular to find the constant.
In this case i'm using the coordinate
(-2, 16)
y = mx + c
16 = -6(-2) + c
16 = 12 + c
c = 4
∴ The equation of the line is y = -6x + 4
The next step is to simply substitude in the x = 8 to the equation to find y.
y = -6(8) + 4
y = -48 + 4
y = -44
When you see a question like this, you need to find the equation of the straight line.
The formular used is y = mx + c
Where
m = slope
c = constant
First find the slope, since it's a straight line, any 2 coordinates can be used.
Now we need to substitude in the slope, and one of the coordinate you used to find the slope, to the formular to find the constant.
In this case i'm using the coordinate
(-2, 16)
y = mx + c
16 = -6(-2) + c
16 = 12 + c
c = 4
∴ The equation of the line is y = -6x + 4
The next step is to simply substitude in the x = 8 to the equation to find y.
y = -6(8) + 4
y = -48 + 4
y = -44