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Why is it with function transformations that sometimes horizontal translations are the opposite of what I think they'd be?

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3 Answers

Take y = x2 and y= (x-4).
 
The parabolic form is y = a(x MINUS h) PLUS k, where (h, k) is the vertex of the parabola opening up or down depending on the sign of a.
 
So the vertex is at (0,0) for y = x2 since y = 1(x-0)2 + 0.
 
Now, for y = (x-4)2, note that the vertex (which is essentially the "handle" for the parabola) is now at
(+4, 0).
 
So despite the (x MINUS 4) being squared, the x coordinate of the vertex was shifted from 0 to 4, which is a shift to the right of 4.
 
 
 
Now consider y=(x+4)2.   
 
That's really y = (x - (-4))2 + 0.   So the vertex was moved 4 units to the left from (0, 0) to (-4, 0).
 
 
Hope this helps.
For quadratic functions such as this one, you should complete the square to see how the parabola is shifted and reflected. In your case,
 
f(x) = –x²– 4x+ 5 = -(x+2)² + 9
 
The +9 tells us the parabola is shifted up (positive y-direction) by 9 units relative to the origin, the +2 tells us it is shifted left by 2 units (negative x-direction), so the vertex is at (-2,9). The negative sign tells us it opens down (in the negative y-direction).
 
More generally, if you change a function from f(x) to f(x+A) for any positive constant A, its graph is shifted A units to the left. f(x-A) shifts the graph A units to the right. If you change f(x) to f(x)+A, it shifts it A units up, and f(x)-A is shifted A units down.
Y=x^2 is centered around origin.  
 
Assuming your coordinates are traditional (+ to the right and - to the left of the origin)
 
Analyze y=x^2 - 4x:
 
To bring y back to the origin, we have to add (move 4x units to the right).  
 
This means y moved -4x units which is 4x to the left.