HELPPPP PLSSSSSSSS
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![HELPPPP PLSSSSSSSS class=](https://us-static.z-dn.net/files/d8a/3b4862098c291a358bd1479155159b86.png)
Answer:
Translation of 3 units to the left.
Vertical stretch by a factor of 2.
Translation of 5 units down.
Step-by-step explanation:
Transformations
For a > 0
[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]
[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]
[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]
Parent function:
[tex]y=x^2[/tex]
Translate 3 units left
Add 3 to the variable of the function
[tex]\implies y=(x+3)^2[/tex]
Stretch vertically by a factor of 2
Multiply the whole function by 2:
[tex]\implies y=2(x+3)^2[/tex]
Translate 5 units down
Subtract 5 from the whole function:
[tex]\implies y=2(x+3)^2-5[/tex]
Please see the attached graphs for the final transformed function (as well as the graphed steps).
Answer:
a) vertical expansion by a factor of 2; translation 3 units left and 5 units down
b) see attached
Step-by-step explanation:
Describing transformations is all about matching patterns. The elements of the transformed function are matched with the elements of a transformation.
A function is scaled vertically by multiplying each function value by some scale factor. In generic terms, the function f(x) is scaled vertically by the factor 'c' in this way:
If we want the function f(x) = x² scaled vertically by a factor of 2, then we have
f(x) = x² . . . . . . . original function
2·f(x) = 2x² . . . . scaled vertically by a factor of 2
On a graph, each point is vertically twice as far vertically from some reference point (the vertex, for example) as it is in the original function graph.
A function is translated to the right by 'h' units when x is replaced by (x -h).
If we want the function f(x) = x² translated right by 3 units, we will have ...
f(x) = x² . . . . . . . . . . . original function
f(x -3) = (x -3)² . . . . . .translated right 3 units
Note that translation left by 3 units would give ...
f(x -(-3)) = f(x +3) = (x +3)² . . . . translated left 3 units
On a graph, each point of the left-translated function is 3 units left of where it was on the original function graph.
A function is translated upward by 'k' units when k is added to the function value.
The value of k will be negative for a translation downward.
If we want the function f(x) = x² translated down by 5 units, we will have ...
f(x) = x² . . . . . . . . . . . original function
f(x) = x² -5 . . . . . . . . .translated down 5 units
Using all of these transformations at once, we have ...
f(x) = x² . . . . . . . . . . . . . . . . . original function
c·f(x -h) +k = c·(x -h)² +k . . . scaled by 'c', translated h right and k up
Compare this to the given function:
y = 2(x +3)² -5
and we can see that ...
This is the pattern matching that is described at the beginning.
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When graphing a transformed function, it is often useful to start with a distinctive feature and work from there. The vertex of a parabola is one such distinctive feature.
The transformations move the vertex 3 units left and 5 units down from its original position at (0, 0). The location of the vertex on the transformed function graph will be at (x, y) = (-3, -5).
The graph of the parent function parabola (y= x²) goes up from the vertex by the square of the number of units right or left. That is, 1 unit right or left of the vertex, the graph is 1 unit above the vertex. 2 units right or left, the graph is 2² = 4 units above the vertex.
The scaled graph will have these vertical distances multiplied by 2:
The graph of the transformed function is shown in blue in the attachment.
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Additional comment
The vertical scale factor 'c' may have any non-zero value, positive or negative, greater than 1 or less than 1. When the magnitude is less than 1, the scaling is a compression, rather than an expansion. When the sign is negative, the graph is also reflected across the x-axis, before everything else.