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Mathematics 10 Online
OpenStudy (anonymous):

The graph of f (in blue) is translated a whole number of units horizontally and vertically to obtain the graph of h (in red). The function f is defined by f(x)= sqrt x Write down the expression for h(x)

OpenStudy (anonymous):

OpenStudy (whpalmer4):

\[y = f(x)+a\]is the graph of \[y = f(x)\]shifted vertically by \(a\) units \[y = f(x-a)\] is the graph of \[y = f(x)\]shifted horizontally by \(a\) units

OpenStudy (primeralph):

That's not right. It's -4 and it's in the square root,

OpenStudy (anonymous):

ive been working on this since yesterday and it still isn't any clearer. ive also been reading other pieces of information on transformations.

OpenStudy (anonymous):

can someone give me the solution and tell me why it works? I have like 12 more of these.

ganeshie8 (ganeshie8):

look at the graph, for definiteness, pick one point : the thick solid point.

ganeshie8 (ganeshie8):

on blue graph, it is at (0, 0) on red graph, it is at (4, 2) right ?

ganeshie8 (ganeshie8):

that means the point (0, 0) shifted to the point (4, 2) : (0, 0) ---> (4, 2)

ganeshie8 (ganeshie8):

it seems, x values are shifting "right" by 4 units, and y values are shifting "up" by 2 units

OpenStudy (anonymous):

would it be: \[f(x)=\sqrt{x-4}+2\] @ganeshie8

OpenStudy (whpalmer4):

It would be. Just like I advised you initially: \[f(x-4) + 2\]to shift horizontally by 4 and vertically by 2

OpenStudy (anonymous):

what happens to the radical?

OpenStudy (anonymous):

never mind I get it it goes under the radical. thank you..

OpenStudy (whpalmer4):

The radical is still there. \[f(x) = \sqrt{x}\]\[f(x)+a = \sqrt{x}+a\]\[f(x-a) = \sqrt{x-a}\]

OpenStudy (whpalmer4):

and \[f(x-a)+b = \sqrt{x-a}+b\]

OpenStudy (whpalmer4):

think of \[f(x) = \sqrt{x}\]as a recipe. It means "whenever you see the expression \(f(expression)\), replace it with \(\sqrt{expression}\)"

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