Geometry › How to find transformation for an analytic geometry equation
Regular pentagons have lines of symmetry through each vertex and the center of the opposite side, meaning the y-axis forms a line of symmetry in this instance. Therefore, point P is negative b units in the x-direction, and c units in the y-direction. It is a reflection of point (b,c) across the y-axis.
Line m passes through the points (–4, 3) and (2, –6). If line q is generated by reflecting m across the line y = x, then which of the following represents the equation of q?
3x + 2y = 6
3x + 2y = 18
2x + 3y = –6
2x + 3y = 6
–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.
Let f(x) = x3 – 2x2 + x +4. If g(x) is obtained by reflecting f(x) across the y-axis, then which of the following is equal to g(x)?
–x3 – 2x2 – x – 4
–x3 + 2x2 – x + 4
x3 + 2x2 + x + 4
–x3 – 2x2 – x + 4
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.
Let . If we let
equal
when it is flipped across the y-axis, what is the equation for
?
When a function is flipped across the y-axis, the resulting function
is equal to
. Therefore, to find our
, we must substitute in
for every
is our equation:
Our final answer is therefore
The graphs of and
are shown above. Which equation best describes the relationship between
and
?
Let f(x) = -2x2 + 3x - 5. If g(x) represents f(x) after it has been shifted to the left by three units, and then shifted down by four, which of the following is equal to g(x)?
-2x^2 + 3x - 12
-2x^2 + 15x - 36
-2x^2 + 31x - 124
-2x^2 - 9x - 18
-2x^2 - 9
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.
If the point (6, 7) is reflected over the line and then over the x-axis, what is the resulting coordinate?
(7, –6)
(6, –7)
(7, 6)
(6, 7)
(–6, –7)
A reflection over the line involves a switching of the coordinates to get us (7, 6). A reflection over the x-axis involves a negation of the y-coordinate. Thus the resulting point is (7, –6).
Let f(x) be a function. Which of the following represents f(x) after it has been reflected across the x-axis, then shifted to the left by four units, and then shifted down by five units?
–f(x – 4) – 5
–f(x + 4) – 5
f(x + 4) – 5
–f(x – 5) – 4
–f(x + 5) – 4
f(x) undergoes a series of three transformations. The first transformation is the reflection of f(x) across the x-axis. This kind of transformation takes all of the negative values and makes them positive, and all of the positive values and makes them negative. This can be represented by multiplying f(x) by –1. Thus, –f(x) represents f(x) after it is reflected across the x-axis.
Next, the function is shifted to the left by four. In general, if g(x) is a function, then g(x – h) represents a shift by h units. If h is positive, then the shift is to the right, and if h is negative, then the shift is to the left. In order, to shift the function to the left by four, we would need to let h = –4. Thus, after –f(x) is shifted to the left by four, we can write this as –f(x – (–4)) = –f(x + 4).
The final transformation requires shifting the function down by 5. In general, if g(x) is a function, then g(x) + h represents a shift upward if h is positive and a shift downward if h is negative. Thus, a downward shift of 5 to the function –f(x + 4) would be represented as –f(x + 4) – 5.
The three transformations of f(x) can be represented as –f(x + 4) – 5.
The answer is –f(x + 4) – 5.
Explain how the below function translates:
5 units left, 7 units down
5 units right, 7 units down
5 units up, 7 units left
5 units down, 7 units right
When estimating the translations for a quadratic function we must remember what vertex form for a parabola looks like:
In order to end up with:
We must have the below to end up with a positive 5:
This is what gives us the translation left 5 spaces and down 7 spaces.
How is different from
?
The slope of is steeper than the slope of
.
The graph of is dilated compared to the graph of
.
The graph of is shifted up three units along the y-axis from the graph of
.
The graph of is shifted to the right three units along the x-axis from the graph of
.
The standard form of a linear equation is Here, we are given two equations,
and
, which differ only in their
terms. In other words, these functions differ only in their slope.
has a larger slope than does
, so
is steeper.