Answer:

Step-by-step explanation:
Write down what the problem is saying;

Distribute;

Simplify,

Inverse operations,

Answer:
translation 2 units left
; reflection across the y-axis
Step-by-step explanation:
The y-coordinates of the points do not change from the pre-image to the image. This means there is no translation down (this would add or subtract to the y-coordinates) and no reflection across the x-axis (this would negate the y-coordinates).
This leaves us with a translation 2 units left and a reflection across the y-axis.
The translation 2 units left adds 2 to the x-coordinates, and the reflection across the y-axis negates the x-coordinates. If we add 2 first, the coordinates would be (-4+2, 6) = (-2, 6); (-2+2, 2) = (0, 2); and (-6+2, 2) = (-4, 2).
Negating each of these would give us (2, 6); (0, 2); and (4, 2). These are the desired image coordinates.
Answer:
(-5, 4]
Step-by-step explanation:
Range is the set of possible outputs for a function within the given domain. It is usually represented in terms of minimum and max values
In the graph, the minimum value is -5 (not including this value) and this occurs at x = 3; the max value is 4 at x = 0
So range is -5 < x ≤ 4
which in interval notation is
(-5, 4]
Answer: y = -2x - 7
Step-by-step explanation:
The equation for a perpenducular line has a slope that is the negative inverse of the reference line, in this case x - 2y = 6.
Let's put this equation into standard slope-intercept form, y = mx+b, where m is the slope and b the y-intercept (the value of y when x is 0).
x - 2y = 6
- 2y = 6 - x
y = (1/2) x -3
The slope is (1/2), so a perpendicular line will have a slope of -2.
y = -2x + b
We can find b by using the given point ((-3,-1), a known solution):
-1 = -2*(-3) + b
-1 = 6 + b
b = -7
The perpendicular equation is y = -2x - 7
y = -2x