A graph has equation y=2x^2-4. What is the equation of the new graph when it is translated 2 units in the positive x direction
2 units to the right is \[f(x-2)\] \[y=2(x-2)^{2}-4\]
these can be opened to \[y=2x^2-8x+4\]
these can be seen as the TP(0;-4) is now (2;-4)
do any of u mind explaining?:D
translating a graph can be done in two ways >shifting it along the y-axis which affects \[f(x)\] if a graph is moved 2 units up \[f(x)+2\] \[y=2x^2-4+2\] and two units down \[f(x)-2\] \[y=2x^2-4-2\] >but when pelleting along the x-axis it is always oppsite ie.2 units right(the positive x direction) affects x\[f(x-2)=2(x-2)^2-4\] ie.2 units left(the positive x direction) affects x\[f(x+2)=2(x+2)^2-4\]
why is it f(x-2)?
\[f(x-a)\]shifts the graph of \(f(x)\) by \(a\) units to the right why? consider some random graph...|dw:1341710473045:dw|at x=a I have marked the point now what about the graph of \(f(x-a)\) ? well, if we change our graph from f(x) to f(x-a) then the point f(a) becomes f(a-a)=f(0) so the point that was at x=a is now f(0) instead of f(a) let's see what that looks like...
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