Primarygun Posted May 11, 2005 Posted May 11, 2005 I am confused of shifting a linear equation. Let f(x)=ax+b And g(x) is identical to f(x+2)+5 For example, we create a specific condition, g(x)=f(x) and (1,2) is a point on f(x) [Does this implies that (1,2) is also a point on g(x)?] Next step is to find f(x): By using the given conditions, f(x)= -5x/2+9/2 The contradiction appears: g(x)=f(x+2)+5 That's mean shifting the whole curve of f(x) to left parallel to x-axis by 2 units, then by shifting it upwards by 5 units, we get g(x). My answer to the previous question ( typed in bold ) is yes but I am not certain with my answer. If I am correct, then the point hasn't moved away. However, it's clear to know that the shifting must move the point upward DUE TO A VECTOR NATURE. My contradiction is here, anyone helps me solve it?
uncool Posted May 11, 2005 Posted May 11, 2005 g(x) = f(x+2) + 5 = -5(x+2)/2 + 9/2 = -5x/2 -5 + 9/2 + 5 = -5x/2 + 9/2 It is because you are moving the line in the same direction as the line is going - the slope is -5/2, or 5 units up over -2 units to the left - which is what you said. However, (1,2) goes to (3,7). -Uncool-
Primarygun Posted May 12, 2005 Author Posted May 12, 2005 Thanks for reply first. How is the point (3,7) archieved? g(1)=f(1+2)+5 Isn't the variable,i.e., the x-coord. is 1 ?
Primarygun Posted May 12, 2005 Author Posted May 12, 2005 Once the function f(x) is changed to g(x)=f(x+2)+5, that means it shifts left by 2 and upwards by 5. Obviously, the new point is not the original point as it has moved upward along the original line. This is a general case for all linear equations. However, as I mentioned, if I let f(x)=g(x), moreover, (1,2) lies on the equation,...........then f(1)=2 ---> g(1)=2 That means the point didn't move. How to explain the italic sentence? My solution: The line is shifted 2 units leftwards and then 2 units rightwards for that specific equation.
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