When a point is reflected across the line y = x, the x and y coordinates are switched. In other words, the point (a, b) reflected across the line y = x would be (b, a).

Thus, if line m is reflected across the line y = x, the points that it passes through will be reflected across the line y = x. As a result, since m passes through (–4, 3) and (2, –6), when m is reflected across y = x, the points it will pass through become (3, –4) and (–6, 2).

Because line q is a reflection of line m across y = x, q must pass through the points (3, –4) and (–6, 2). We know two points on q, so if we determine the slope of q, we can then use the point-slope formula to find the equation of q.

First, let's find the slope between (3, –4) and (–6, 2) using the formula for slope between the points (x1, y1) and (x2, y2).

slope = (2 – (–4))/(–6 –3)

= 6/–9 = –2/3

Next, we can use the point-slope formula to find the equation for q.

y – y1 = slope(x – x1)

y – 2 = (–2/3)(x – (–6))

Multiply both sides by 3.

3(y – 2) = –2(x + 6)

3y – 6 = –2x – 12

Add 2x to both sides.

2x + 3y – 6 = –12

Add six to both sides.

2x + 3y = –6

The answer is 2x + 3y = –6.

When a point is reflected across the line y = x, the x and y coordinates are switched. In other words, the point (a, b) reflected across the line y = x would be (b, a).

Thus, if line m is reflected across the line y = x, the points that it passes through will be reflected across the line y = x. As a result, since m passes through (–4, 3) and (2, –6), when m is reflected across y = x, the points it will pass through become (3, –4) and (–6, 2).

Because line q is a reflection of line m across y = x, q must pass through the points (3, –4) and (–6, 2). We know two points on q, so if we determine the slope of q, we can then use the point-slope formula to find the equation of q.

First, let's find the slope between (3, –4) and (–6, 2) using the formula for slope between the points (x1, y1) and (x2, y2).

slope = (2 – (–4))/(–6 –3)

= 6/–9 = –2/3

Next, we can use the point-slope formula to find the equation for q.

y – y1 = slope(x – x1)

y – 2 = (–2/3)(x – (–6))

Multiply both sides by 3.

3(y – 2) = –2(x + 6)

3y – 6 = –2x – 12

Add 2x to both sides.

2x + 3y – 6 = –12

Add six to both sides.

2x + 3y = –6

The answer is 2x + 3y = –6.

If Jenny moves Bobby's circle up 2 units and to the right 1 unit, then the center of her circle is (3, 7). The radius remains 10.

The general equation for a circle with center (h, k) and radius r is given by

For Jenny's circle, (h, k) = (3, 7) and r=10.

Substituting these values into the general equation gives us

If Jenny moves Bobby's circle up 2 units and to the right 1 unit, then the center of her circle is (3, 7). The radius remains 10.

The general equation for a circle with center (h, k) and radius r is given by

For Jenny's circle, (h, k) = (3, 7) and r=10.

Substituting these values into the general equation gives us

In order to reflect a function across the y-axis, all of the x-coordinates of every point on that function must be multiplied by negative one. However, the y-values of each point on the function will not change. Thus, we can represent the reflection of f(x) across the y-axis as f(-x). The figure below shows a generic function (not f(x) given in the problem) that has been reflected across the y-axis, in order to offer a better visual understanding.

Therefore, g(x) = f(–x).

f(x) = x3 – 2x2 + x – 4

g(x) = f(–x) = (–x)3 – 2(–x)2 + (–x) + 4

g(x) = (–1)3x3 –2(–1)2x2 – x + 4

g(x) = –x3 –2x2 –x + 4.

The answer is –x3 –2x2 –x + 4.

In order to reflect a function across the y-axis, all of the x-coordinates of every point on that function must be multiplied by negative one. However, the y-values of each point on the function will not change. Thus, we can represent the reflection of f(x) across the y-axis as f(-x). The figure below shows a generic function (not f(x) given in the problem) that has been reflected across the y-axis, in order to offer a better visual understanding.

Therefore, g(x) = f(–x).

f(x) = x3 – 2x2 + x – 4

g(x) = f(–x) = (–x)3 – 2(–x)2 + (–x) + 4

g(x) = (–1)3x3 –2(–1)2x2 – x + 4

g(x) = –x3 –2x2 –x + 4.

The answer is –x3 –2x2 –x + 4.

The general formula for a circle with center (h,k) and radius r is .

The center of the original circle, therefore, is (2, -4).

If we move the circle to the left 2 spaces and down 3 spaces, then the center of the new circle is given by or .

The general formula for a circle with center (h,k) and radius r is .

The center of the original circle, therefore, is (2, -4).

If we move the circle to the left 2 spaces and down 3 spaces, then the center of the new circle is given by or .

We are told that g(x) is found by taking f(x) and shifting it to the left by three and then down by four. This means that we can represent g(x) as follows:

g(x) = f(x + 3) - 4

Remember that the function f(x + 3) represents f(x) after it has been shifted to the LEFT by three, whereas f(x - 3) represents f(x) after being shifted to the RIGHT by three.

f(x) = -2x2 + 3x - 5

g(x) = f(x + 3) - 4 = \[-2(x+3)2 + 3(x+3) - 5\] - 4

g(x) = -2(x2 + 6x + 9) + 3x + 9 - 5 - 4

g(x) = -2x2 -12x -18 + 3x + 9 - 5 - 4

g(x) = -2x2 - 9x - 18 + 9 - 5 - 4

g(x) = -2x2 - 9x - 18

The answer is -2x2 - 9x - 18.