Generalizing: the image of x by the first function is the same as the image of x+3 by the second function.
If we write the first function as f(x), then the second function is f(x-3).
This means that the graph of y=f(x-3) arises when we move the graph of y = f(x)
three units to the right.
Exercise : plot the two functions and notice the translation.
The graph of y=f(x-a) arises when we move the graph of y = f(x) exactly a units to the right.
We choose a arbitrary fixed vector v(a,b).
We shift the graph of y=f(x) by the vector v.
First we move the graph a units to the right. Then we move the graph b units upwards.
The new equation is y = f(x-a) + b
Conclusion:
Is we submit the graph of y= f(x) to a translation by the vector v(a,b), then the image
is the function with equation y = f(x-a) + b.
We start with the function with equation y = 5 cos(2x-6) + 4
We want a suitable composition of elementary transformations such that the
equation simplifies to y = cos(x).
First, we translate the graph 4 units downwards.
The equation becomes y = 5 cos(2x-6) <=> y = 5 cos(2 (x-3))
Now, we translate the graph 3 units to the left.
The equation becomes y = 5 cos(2x)
We multiply all images with factor 1/5.
The equation becomes y = cos(2x)
Finally, we stretch the graph away from the y-axis with factor 2.
The equation becomes y = cos(x).
We translate the parabola with equation y = 2x^{2} to the right such that the vertex
is on the line y = 2x - 5
Find the equation of the translated parabola.
The intersection of the line with the x-axis is point (5/2 , 0).
We have to translate the parabola to the right by a distance 5/2.
The equation of translated parabola is
y = (x - 5/2)^{2}
<=>
y = x^{2} - 5x + 25/4
We translate the parabola y = x^{2} upwards until the line y = x is tangent to the parabola.
Find the equation of the translated parabola.
We translate the parabola a units upwards. Then , the equation is y = x^{2} +a.
The line y = x is tangent to the parabola y = x^{2} +a
<=>
There is only one intersection point of y = x and y = x^{2} +a
<=>
The following system has two equal solutions
/ y = x
\ y = x^{2} +a
<=>
x^{2} - x + a = 0 has two equal solutions
<=>
1 - 4a = 0
<=>
a = 1/4
The translated parabola is y = x^{2} + 1/4
The parabola y = x^{2} is compressed towards the y-axis. This compressed parabola
intersects the line y=x in point S(1/4,1/4).
Find the equation of the transformed parabola.
The parabola y = x^{2} is compressed towards the y-axis with a factor a.
Then the equation of the parabola is y = (ax)^{2}.
S is on the parabola if and only if 1/4 = a^{2} .(1/16).
So, a =2.
The transformed parabola has an equation y = 4 x^{2}
The point S(1,1) is a point of the parabola y=x^{2}.
We stretch the parabola away from the y-axis until the distance from O(0,0)
to S is doubled relative to the original distance.
Find the equation of the stretched parabola.
We stretch the parabola away from the y-axis with factor a.
Then the equation of the parabola is y=(x/a)^{2} or also y=x^{2}/a^{2}.
Now, point S has (a,1) as coordinates and the distance |OS|^{2} is 1 + a^{2}.
The original value of |OS|^{2} was 2. Thus, in the stretched position, |OS|^{2} must be 8.
1 + a^{2} = 8 <=> a^{2} = 7 .
The stretched parabola has y = x^{2}/7 as equation.
We compress the graph of y = 3 sin(2 x - 1) towards the y-axis with a suitable factor a
such that the period of the new graph is equal to 1.
Find the value of a and the equation of the new graph.
After the compression with factor a, the new equation is y = 3 sin(2a x - 1).
The period of this function is (2 pi)/(2 a).
The period is 1 if and only if a = pi.
Then the equation is y = 3 sin(2pi x - 1).
We translate the graph of y = 2 cos^{2}(2x) downwards by a suitable distance v
such that the translated graph is symmetric relative to the x-axis.
Find v and the equation of the translated graph.
By the Carnot-formulas, we know that 2 cos^{2}(2x) = 1 + cos(4x).
Then, the given equation is y = cos(4x) + 1.
We translate the graph 1 unit downwards.
The equation of the translated graph is y = cos(4x) and v = 1.
We translate the graph of y = 3 cos(2 pi x + 4) to the left
by a positive distance a.
Find the minimum value for a such that the origin O(0,0) is on the
translated graph.
The equation of the translated graph is
y = 3 cos(2 pi (x+a) + 4).
<=>
y = 3 cos(2 pi x + 4 + 2 pi a)
O(0,0) is on the translated graph.
<=>
0 = 3 cos(4 + 2 pi a)
<=>
cos(4 + 2 pi a) = 0
<=>
4 + 2 pi a = pi/2 + 2 k pi of 4 + 2 pi a = - pi/2 + 2 k pi
<=>
...
<=>
a = -0.3866 + k of a = -0.8866 + k
The minimum positive value for a is 0.11338
Tip: Check this result with a plotter
We translate the graph of y = cos(x+2) + cos(x-2) upwards
by a positive distance b such that the translated graph is tangent to the x-axis.
Find the value of b.
The translated graph has as equation
y = cos(x+2) + cos(x-2) + b
<=>
y = 2 cos(x) cos(2) + b
<=>
y = 2 cos(2) cos(x) + b
Since 2 cos(2) is negative, the value of b is -2 cos(2).
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