Reflection across the x-axis M(0,-1) S(3,2) Q(5,-3)
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F(-1)= -3 is (-1,-3) and f(2) = 6 is (2,6) where f(x) = y
y=mx + b is the slope-intercept form whereas m equals the slope (rate of change) and b equals the y-intercept (initial amount/what y is when x is 0.)
First, we need to find the slope between the two points (-1,-3) and (2,6). To find the slope we could use one of it's formulas
.
1. (-1,-3)
2. (2,6)
→ 
→ 
The slope is 3 (). Thusly, y = 3x + b
To find out the y-intercept, we can reverse the slope. [Note: This is in 
where rise is 'y' and run is 'x'. Reversed would be 
]. Take the second ordered pair and use our reversed slope on it until we get 0 for x.
(2, 6) ⇒ (2 - 1, 6 -3) ⇒ (1, 3) ⇒ (0,0)
Y-intercept is 0. Therefore, y= 3x + 0 [NOTE: y = f(x), so if you want it in function notation form it's just f(x) = 3x + 0.]
y=mx + b is the slope-intercept form whereas m equals the slope (rate of change) and b equals the y-intercept (initial amount/what y is when x is 0.)
First, we need to find the slope between the two points (-1,-3) and (2,6). To find the slope we could use one of it's formulas
1. (-1,-3)
2. (2,6)
The slope is 3 (
To find out the y-intercept, we can reverse the slope. [Note: This
(2, 6) ⇒ (2 - 1, 6 -3) ⇒ (1, 3) ⇒ (0,0)
Y-intercept is 0. Therefore, y= 3x + 0 [NOTE: y = f(x), so if you want it in function notation form it's just f(x) = 3x + 0.]