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I am having trouble shifting graphs when you need to define a function as y1= then you need to subtract or add to move it right or left. Lets say you have 

y1=√(1-x^2)

how do you then shift that to the right and left? When you do 

y2=y1-2

that shifts it up two. How would you shift this over two? you need to define y2 in terms of y1

user2330624
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  • The graph of $y = f(x - h) + v$ is the graph of $y = f(x)$ shifted to the right by $h$ units (so if $h$ is negative, it has been shifted to the left by $|h|$ units), and shifted up by $v$ units (so if $v$ is negative, it has been shifted down by $|v|$ units). – Michael Albanese Jun 20 '15 at 23:25

2 Answers2

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Remember $y_1$ and $y_2$ are functions; so we can also work with its input.

In order to shift the graph horizontally, say two to the right, we need the value of the original function, $y_1(x)$, to be the same as the value of the new function two to the right, $y_2(x + 2)$. In other words, we want $$ y_2(x + 2) = y_1(x) $$ So a simple substitution gives $$ y_2(x) = y_1(x - 2) $$ For your example in particular, we have $y_2(x) = y_1(x - 2) = \sqrt{1 - (x - 2)^2}$. You can easily generalize this to arbitrary horizontal shifts to the left or right.

Pratyush Sarkar
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Suppose you have a function $y=f(x)$. If you intend to move the graph by $a$ units towards the right graph the function $y=f(x-a)$. Notice that if $(x_0,y_0)$ is any point on the previous graph ($y=f(x)$), then $(x_0+a,y_0)$ is a point on the new graph ($y=f(x-a)$).

QED
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