So I have plato summer school and some of the units are just crazy confusing. HELP!!!!

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The graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.

So, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2).

It is apparent that I have fallen victim to one of the classic blunders – the most famous of which is: Never get involved in a land war in Asia – but only slightly less well known is this: Never attempt to answer an Algebra question if travel worn whilst sleep deprived!

Ahah haha, ahah haha, aha... ugh. :(

Mark and Elvis are correct, as is obvious to a well-rested mind, this an example of the Horizontal Translation of a function.

However, as neither the response of Mark nor Elvis serves to elucidate how we arrive at the solution, as much as state the solution, I will attempt to make amends by providing a more thorough explanation:

As accurately stated by Mark and Elvis, f(x-2) is – and always will be – the graph of f(x) shifted by two units to the right along the x-axis (a Horizontal Shift), just as the graph of f(x)+2 is – and always will be – the graph of f(x) shifted upwards by two units along the y-axis (a Vertical Shift).

It is important to note that Horizontal and Vertical Shifts are but two of the six types of Function Translations, which include Horizontal and Vertical Scaling, and Reflection about the x and y axes (though Reflections are technically special cases of the operation involved in Scaling).

Furthermore, you may notice that if we take our expression for the Vertical Shift, f(x)+2, and restate it as y=f(x)+2, we get something that seems an awful lot like a Linear Equation of the Slope-Intercept form. Indeed, it is an equation with a slope of 1 and a y-intercept of 2. This demonstrates the fundamental fact that all linear equations are, in effect, Translations of the Generic Linear Function f(x).

To understand why this is, it is important to grasp what the notation is actually meant to communicate. The function f(x) is generic because it is the most fundamental form of a linear equation, the Identity y=x. This is made more clear if we restate f(x) more succinctly, if less concisely, as y=f(x). Since the Generic Linear Function (GLF) is an identity, every coordinate pair it produces will also be an Identity, and the graph of the function is a line of slope 1 with its x and y intercepts both located at (0,0).

Our translation of the GLF to f(x-2), or y=f(x-2), shifts the graph to the right by two units, resulting in the need to travel two units further along the x-axis in order to arrive at the corresponding y-coordinate.

This holds true no matter how we translate the GLF.

For example, in your case, the functions f(x)=x-7, f(x)=1/9x+1, and f(x)=2x-16 will all render the coordinate pair (9,2).

f(x)=x-7 translates the GLF by shifting it down the y-axis by seven units.

f(x)=1/9x+1 translates the GLF by reducing it to 1/9 scale (this is effectively what a slope adjustment is), and shifting it up the y-axis by one unit.

f(x)=2x-16 translates the GLF by increasing it to 2 times scale, and shifting it down the y-axis by 16 units.

By translating these functions through f(x-2), we are performing the same translations above, but also shifting the graph by 2 units to the right along the x-axis. This may seem counter-intuitive given that it is the reverse of how Vertical Shifts work, but it makes sense if we consider that what is effectively happening is that the x-coordinate is being increased by 2 in order to render the same y-coordinate.

Unfortunately it is impossible to demonstrate this graphically here, and the editor does not allow for easy formatting when emending a previous response, so I have uploaded graphical examples of the GLF, the f(x-2) translation, the f(x)=x-7 and f(x-2)=7 translations, as well as constrained tables demonstrating everything discussed above.

These can be found in my Wyzant Locker at:

and

As you can see from these graphs and tables, regardless of the translation of the GLF, any translation that produces the coordinate pair (9,2) for f(x) will produce the corresponding coordinate pair (11,2) for f(x-2).

QED

I hope this serves to atone for my previous error, which follows below.

Good afternoon Kiera,

As presented above, the solution to this problem is indeterminable, as there are an infinite number of linear equations for which the point (9,2) could be a solution.

Does the problem come with a graphical representation of f(x) from which the slope, the y-intercept, or both can be determined? If you are able to determine either of these constants, then you can make soundly reasonable assumptions, otherwise the problem is unsolvable.

Ultimately you would need to be able to determine that linear equation f(x) in the form f(x)=mx+b and then solve for

'x-2' to determine how this adjusts the graph.

Hope this helps.

-Dennis

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