Stanton D. answered • 02/22/22
Tutor to Pique Your Sciences Interest
Hi Alberto M.,
What I'll do is help you on how to think about this problem.
First of all, the question as it is on Wyzant, for the parent function being y=abx, is clearly trash. Fortunately, you did include a web reference to the original. But even without looking at that, I can confidently state that that should have been written as: y = a + bx . That is the usual "slope intercept" form of a function, which simply means that the given equation immediately tells you that the y-intercept (where the function graph crosses the y-axis) is at the value "a", and the slope of the straight line (which is the graph of the function) has the value "b". Are you OK with that so far, I hope?
Now, why don't you draw such a graph, for example with y-intercept at 3, and slope of 3. So the graph will also pass through the point (-1, 0) as well as the point (0, 3), and be a straight line. The equation for that line would be: y= 3 + 3x
So you want a transformed graph, displaced by the indicated amounts.
Here is what you are going to do and think. For the displacement left by 2 units: you are going to imagine the graph you drew, moved left by 2 units. Now, what would you have to do to each point of that new graph so that it would match the graph you originally drew? You would have to move each point back to the right by 2 units. To do that, you would take the new x-value of some point, and add 2 units to it; that would move it to the right, wouldn't it. So we will write the equation for the new graph, but move each point to the right, by substituting (x+2) whereever we had just an x in the equation.
So, so far, we have y = a + b(x+2) .
Now, we only moved the "new graph" back to right, to match the original graph. But there was also a movement downwards specified, too. So, to get the "new" graph back to where the original one was, we have to move it up by 5 units. Since we are moving the graph in the y-direction, we add the 5 units to the original y: (y+5) = a + b(x+2) . So look at that equation, as I walk you through your mental checklist: This is the graph of the moved function. But if I move each new y up by 5 (by adding 5) and move each new x right by 2 (by adding 2), I will have my original graph back exactly as I drew it. Yes, that is correct!
All that remains is to simplify the new-graph equation into standard form:
y = (a-5+2b) + bx . Note that the slope (b) of the new graph didn't change from that of the original, but the y-intercept certainly did. For your particular graph I suggested above, would work out as y = (3-5+6) + 3x = 4 + 3x . You might want to graph that, and confirm that it is moved correctly as stated. Of course, if you look closely, you may realize that there are other movements of your original line that would have given the same new graph -- in this case, movement up by 1 unit, and no x movement, would have done the same. But that particular result would only be true for a particular combination of an original graph, and a specified movement.
I hope this technique, and way of thinking about transformations, helps you with these types of problems.
--Cheers, --Mr. d.
