Transformations: f(x) + k, k·f(x), f(x + k)

In this lesson you will: Predict the effect of f(x)+k, kf(x) and f(x+k) on the graph of f(x)=x^2; Give the vertex of a transformed parabola without plotting a single point; Translate between a sentence, function notation and a rule on ordered pairs.

Outside the function, the change is vertical

Start from f(x)=x^2. Anything you do after the squaring acts on the output, and outputs are heights.

f(x)+k adds k to every height, sliding the whole graph up when k>0 and down when k<0. kf(x) multiplies every height: a factor larger than 1 stretches the graph vertically and makes it look narrower, a factor between 0 and 1 compresses it and makes it look wider, and a negative factor flips it over the horizontal axis.

Inside the function, the change is horizontal — and backwards

f(x+3) means: before squaring, add 3 to the input. The new graph therefore does at x=-3 what the old one did at 0, because -3+3=0. Everything slides three units left.

That is the opposite of what the plus sign suggests, and it is the reason this is the most-missed idea in the chapter. The reliable way to see it is to ask which input makes the inside of the parentheses zero.

Three ways to say the same move

A transformation can be described in a sentence, in function notation, or as a rule on ordered pairs. Being fluent in all three is most of what this standard asks for.

For example, g(x)=-(x+2)^2+7 is: shift left 2, reflect over the horizontal axis, then shift up 7. Its vertex is (-2,7) and it opens downward, so 7 is a maximum value. Check one point: g(0)=-(2)^2+7=3, so the y-intercept is (0,3).

Common mistake — Reading f(x + 3) as a shift to the right

Asked to describe g(x)=(x+3)^2, a student writes \"the graph of y=x^2 moved 3 units right.\"

The reasoning behind it is honest and almost always the same: a plus sign means add, adding moves you right along a number line, so the graph must move right. Every step of that sentence is true about numbers and false about this graph.

The vertex of y=x^2 is at x=0. The vertex of y=(x+3)^2 is where (x+3)^2 is smallest, which is where x+3=0, namely x=-3. The graph moved left.

The mirror version is just as common: calling g(x)=(x-4)^2 a shift left. It moves right, to x=4.

How to catch it: Set the inside of the parentheses equal to zero and solve. For f(x+3), solving x+3=0 gives x=-3 — a negative number, so the move is to the left. For f(x-4), solving x-4=0 gives x=4, so the move is to the right. One tiny equation, no memory required.