Respuesta :
we would like to begin looking at the transformations of the graphs of functions. In particular, we will use our familiarity with quadratic equations;
with a≠0, where we will be concerned with three general types of transformations in the variables x and y. Let us first look specifically at the basic monic quadratic equation for a parabola with vertex at the origin, (0,0): y = x². Its graph is given below.
We will consider horizontal translations, horizontal scaling, vertical translations and vertical scaling first.
Horizontal translations affect the domain on the function we are graphing. Recall that the domain is the set of all values that we can put in for x in the function without breaking a rule of algebra, such as division by 0, or taking the logarithm of a negative number. If we translate by some positive real number a, then our parabolas equation is changed "in the parentheses". When we say "in the parentheses" in this context we are referring to the notation: f(x) = x²; we will mean that we are changing the value in the parentheses. We denote a horizontal translation as follows: y = (x-a)². Let's look at how this affects our graph for a=2:
Notice how the graph moved to the right by 2 units. These are the rules we will follow: We will always write our equations in the form "x-something" and resulting horizontal translation is in the direction of the sign of the something and a number of units equal to its absolute value. The two possibilities are that we have either y = (x-a)², or y = (x+a)² = (x-(-a))². Now on to horizontal scaling.
Horizontal scaling involves multiplying inside the parentheses by a nonzero constant b ≠ 0. If b>1, the values of y associated with a given x in the domain occur in "1/b time" in the absence of a horizontal translation, and if b<1, the values of y associated to a given x in the domain occur in "b time". Let's consider a horizontal scaling by b = 2 > 1 of our basic parabola. Our function will take on the form: y = (2x)² = 4x². Understanding what we mean by 1/b time. Consider our basic parabola without the scaling: y = x², here, if we set x = 4, then y = 16. On the other hand, if we put x = 2 into our scaled equation, we get 16. For our scaled equation we see that the value y =16 is achieved at x=2, 1/2 = 1/b of that needed in the unscaled equation. Let's look at the graph of y = (2x)².
We will now consider vertical translations and scaling. Such transformations affect the range of the function and happen "outside" the parentheses. Let's begin by considering a vertical translation by d ≠ 0, where when d > 0 our translation move the graph vertically up d units, and when d < 0 it the graph is moved vertically down d units. Let's look again at our basic parabola after such a transformation. Let d = 2, then our equation takes the form y = x² + 2. Notice that the 2 is not inside the parentheses as before, and is not squared as a result. Here is a picture of the graph of this equation.
As the last of our basic transformations we will give a vertical scaling. This differs from the horizontal scaling as follows: In the horizontal scaling the x-value needed to produce a given y-value was scaled. However, in the case of a vertical scaling, the y-value resulting from a given x-value is scaled. A vertical scaling by a nonzero constant d > 0 will "grow" the y-value associated to some x-value in the domain by d if d > 1, and "shrink" it if d < 1. Let's look at the vertical scaling of our basic parabola by c = 2.
Now, how do we insure we get the correct graph? Consider the following possible graphs for the equation y = (5x-3)², where we have translated first in the top graph and scaled first in the bottom graph.
How do we know which is correct. One simple manner to do so is to plug some different values of x into the equation y = (5x-3)² and compare the resulting y-values to each of the graphs. Doing so rules out the top graph, pointing us to the correct graph. How could we have known this without checking different x-values? The correct order is a consequence of the order of operations we learned when we were first learning about multiplication and addition; we will always handle scaling first, then translations.
What of the transformations I alluded to earlier not already covered? The remaining transformations are horizontal and vertical reflections. The former affects the domain, taking the x-value which produces a specific y-value and sending it to its negative. In our equation it is manifested by allowing our b-values from the scaling above to take on negative value. Because it is a type of scaling, it is handled before translations. Let's look at a graph of the horizontal reflection of the parabola with equation: y = (x+2)². It is given by the equation: y = (-x+2)². The purple graph is associated to the former, and the red to the latter. Notice that the horizontal reflection of a graph is across the y-axis.
with a≠0, where we will be concerned with three general types of transformations in the variables x and y. Let us first look specifically at the basic monic quadratic equation for a parabola with vertex at the origin, (0,0): y = x². Its graph is given below.
We will consider horizontal translations, horizontal scaling, vertical translations and vertical scaling first.
Horizontal translations affect the domain on the function we are graphing. Recall that the domain is the set of all values that we can put in for x in the function without breaking a rule of algebra, such as division by 0, or taking the logarithm of a negative number. If we translate by some positive real number a, then our parabolas equation is changed "in the parentheses". When we say "in the parentheses" in this context we are referring to the notation: f(x) = x²; we will mean that we are changing the value in the parentheses. We denote a horizontal translation as follows: y = (x-a)². Let's look at how this affects our graph for a=2:
Notice how the graph moved to the right by 2 units. These are the rules we will follow: We will always write our equations in the form "x-something" and resulting horizontal translation is in the direction of the sign of the something and a number of units equal to its absolute value. The two possibilities are that we have either y = (x-a)², or y = (x+a)² = (x-(-a))². Now on to horizontal scaling.
Horizontal scaling involves multiplying inside the parentheses by a nonzero constant b ≠ 0. If b>1, the values of y associated with a given x in the domain occur in "1/b time" in the absence of a horizontal translation, and if b<1, the values of y associated to a given x in the domain occur in "b time". Let's consider a horizontal scaling by b = 2 > 1 of our basic parabola. Our function will take on the form: y = (2x)² = 4x². Understanding what we mean by 1/b time. Consider our basic parabola without the scaling: y = x², here, if we set x = 4, then y = 16. On the other hand, if we put x = 2 into our scaled equation, we get 16. For our scaled equation we see that the value y =16 is achieved at x=2, 1/2 = 1/b of that needed in the unscaled equation. Let's look at the graph of y = (2x)².
We will now consider vertical translations and scaling. Such transformations affect the range of the function and happen "outside" the parentheses. Let's begin by considering a vertical translation by d ≠ 0, where when d > 0 our translation move the graph vertically up d units, and when d < 0 it the graph is moved vertically down d units. Let's look again at our basic parabola after such a transformation. Let d = 2, then our equation takes the form y = x² + 2. Notice that the 2 is not inside the parentheses as before, and is not squared as a result. Here is a picture of the graph of this equation.
As the last of our basic transformations we will give a vertical scaling. This differs from the horizontal scaling as follows: In the horizontal scaling the x-value needed to produce a given y-value was scaled. However, in the case of a vertical scaling, the y-value resulting from a given x-value is scaled. A vertical scaling by a nonzero constant d > 0 will "grow" the y-value associated to some x-value in the domain by d if d > 1, and "shrink" it if d < 1. Let's look at the vertical scaling of our basic parabola by c = 2.
Now, how do we insure we get the correct graph? Consider the following possible graphs for the equation y = (5x-3)², where we have translated first in the top graph and scaled first in the bottom graph.
How do we know which is correct. One simple manner to do so is to plug some different values of x into the equation y = (5x-3)² and compare the resulting y-values to each of the graphs. Doing so rules out the top graph, pointing us to the correct graph. How could we have known this without checking different x-values? The correct order is a consequence of the order of operations we learned when we were first learning about multiplication and addition; we will always handle scaling first, then translations.
What of the transformations I alluded to earlier not already covered? The remaining transformations are horizontal and vertical reflections. The former affects the domain, taking the x-value which produces a specific y-value and sending it to its negative. In our equation it is manifested by allowing our b-values from the scaling above to take on negative value. Because it is a type of scaling, it is handled before translations. Let's look at a graph of the horizontal reflection of the parabola with equation: y = (x+2)². It is given by the equation: y = (-x+2)². The purple graph is associated to the former, and the red to the latter. Notice that the horizontal reflection of a graph is across the y-axis.
Answer:
The vertex form of each function is [tex]A(x)=x^2-8[/tex], [tex]B(x)=(x+7)^2[/tex], [tex]C(x)=(x-5)^2+4[/tex] and [tex]D(x)=(x-12)^2[/tex].
Step-by-step explanation:
The given function is
[tex]f(x)=x^2[/tex]
The vertex form of a parabola is
[tex]y=a(x-h)^2+k[/tex]
Where, (h,k) is vertex.
If is given that the functions A(x), B(x), C(x) and D(x) are transformed by f(x). So, the value of a is 1 for each function.
The graph of A(x) shifts 8 units down and the vertex of the parabola is (0,-8). Therefore the vertex form of A(x) is
[tex]A(x)=(x-0)^2+(-8)[/tex]
[tex]A(x)=x^2-8[/tex]
The graph of B(x) shifts 7 units left and the vertex of the parabola is (-7,0). Therefore the vertex form of B(x) is
[tex]B(x)=(x-(-7))^2+(0)[/tex]
[tex]B(x)=(x+7)^2[/tex]
The graph of C(x) shifts 5 units right and 4 units up and the vertex of the parabola is (5,4). Therefore the vertex form of C(x) is
[tex]C(x)=(x-5)^2+4[/tex]
The graph of D(x) shifts 12 units right and the vertex of the parabola is (12,0). Therefore the vertex form of D(x) is
[tex]D(x)=(x-(12))^2+(0)[/tex]
[tex]D(x)=(x-12)^2[/tex]
Therefore the vertex form of each function is [tex]A(x)=x^2-8[/tex], [tex]B(x)=(x+7)^2[/tex], [tex]C(x)=(x-5)^2+4[/tex] and [tex]D(x)=(x-12)^2[/tex].
