Identify the effect on a graph of replacing f(x) by f(x) + k, kf(x), f(kx) and f(x + k), find k from graphs, and recognize even and odd functions

Learning goal: Identify the effect on the graph of a function of replacing f(x) by f(x) + k, by kf(x), by f(kx) and by f(x + k) for specific values of k, both positive and negative; find the value of k when the original and transformed graphs are given; experiment with cases using technology and illustrate an explanation of the effects; and recognize even and odd functions from their graphs and from their algebraic expressions. The examples focus on transformations of graphs of quadratic functions, with other function types used to show that the rules are general.

Two course routes. In Algebra 1 and Math 2 (F.BF.3a) the focus is transformations of quadratic graphs by f(x) + k, kf(x) and f(x + k), finding k from graphs and recognizing even and odd functions; f(kx) is excluded there. In Algebra 2 and Math 3 (F.BF.3) all four replacements, including f(kx), apply to any function type. The paragraphs and tasks below on f(kx), and on functions other than quadratics, belong to the Algebra 2 and Math 3 route.

The reference function and the four replacements. Let f be a function whose graph you know, with the quadratic f(x) = x² as the running example: vertex at (0, 0), passing through (1, 1), (2, 4) and (3, 9), symmetric about the vertical axis. Each of the four replacements below builds a new function from f, and each has one geometric effect that holds for every function, not only for x².

Replacing f(x) by f(x) + k adds k to every output. Every point on the graph moves straight up by k when k is positive and straight down by |k| when k is negative. For f(x) = x², the graph of x² + 3 has vertex (0, 3) and passes through (1, 4) and (2, 7); the graph of x² − 4 has vertex (0, −4) and crosses the horizontal axis at x = 2 and x = −2, since 4 − 4 = 0. The shape is unchanged; only the vertical position moves.

Replacing f(x) by kf(x) multiplies every output by k. When k > 1 the graph is stretched vertically away from the horizontal axis; when 0 < k < 1 it is compressed toward that axis; when k is negative it is also reflected across the horizontal axis. For f(x) = x², the graph of 2x² passes through (1, 2) and (2, 8), twice as high as f at every input, so it is narrower. The graph of ½x² passes through (2, 2) and (4, 8), wider. The graph of −x² is the same parabola opening downward, and −x² + 4, which is −f(x) + 4, is that reflected parabola shifted up to vertex (0, 4), passing through (1, 3). Points on the horizontal axis do not move under kf(x), because k · 0 = 0.

Replacing f(x) by f(x + k) shifts the graph horizontally, in the direction opposite to the sign of k. This is the replacement students most often get backwards, so the reasoning matters. The graph of g(x) = f(x + 3) has the value at input x that f had at input x + 3. So g reaches f's vertex value, 0, when x + 3 = 0, that is, at x = −3: the vertex has moved 3 units left, not right, and the graph of (x + 3)² has vertex (−3, 0). Likewise (x − 2)² = f(x − 2) is f(x + k) with k = −2, and its vertex is at x = 2, moved 2 units right, passing through (3, 1) and (4, 4). The rule is: positive k moves the graph left by k, negative k moves it right by |k|, and the reason is that the input must compensate for the added k before f can act.

Replacing f(x) by f(kx) changes the horizontal scale. The graph of f(kx) at input x shows f's value at kx. When k > 1, f's features arrive at smaller inputs, so the graph is compressed horizontally toward the vertical axis by the factor 1/k; when 0 < k < 1 it is stretched horizontally; when k is negative it is also reflected across the vertical axis. For f(x) = x², f(2x) = (2x)² = 4x², which passes through (1, 4) and (1.5, 9): the output 9 that f reached at x = 3 is reached at x = 1.5, half as far out. For f(x) = √x, the graph of √(4x) = 2√x passes through (1, 2) and (4, 4), and the value √4 = 2 that f reached at x = 4 is reached at x = 1. For x², f(kx) looks the same as a vertical stretch, since (2x)² = 4x². The same coincidence happens for every power function centered at the origin, √(4x) = 2√x and (2x)³ = 8x³ among them, which is why the quadratic focus sets f(kx) aside. For such a power function the horizontal effect shows clearly only once the graph has a feature away from the vertical axis: for f(x) = (x − 2)², the graph of f(2x) = (2x − 2)² has its vertex at x = 1 instead of x = 2. For many other functions the coincidence does not happen: for f(x) = 2ˣ, f(2x) = 4ˣ is not any vertical stretch of 2ˣ, because 4ˣ ÷ 2ˣ = 2ˣ is not a constant.

Finding k from graphs. Given the graph of f and the graph of one transformation, identify which replacement was used from the kind of change, then read k from a matching pair of points. If the vertex of a parabola has moved from (0, 0) to (0, −4) with the shape unchanged, the replacement is f(x) + k with k = −4. If the vertex has moved to (−4, 0), the replacement is f(x + k) with k = 4, because the graph moved left. If the vertex is still at (0, 0) but the graph passes through (1, 3) instead of (1, 1), the replacement is kf(x) with k = 3, found by dividing the new output by the old at the same input. If the graph of √x has been changed so that it passes through (1, 3), and the change is horizontal, then √(k · 1) = 3 gives k = 9, and the check √(9 · 4) = 6 confirms that the transformed graph passes through (4, 6). Both positive and negative values of k must be considered: a parabola with vertex (0, 0) that opens downward through (1, −2) is kf(x) with k = −2, a reflection across the horizontal axis combined with a vertical stretch by the factor 2.

Experimenting with technology. A graphing tool with a slider for k is the fastest way to see these effects and to illustrate an explanation. Enter y = x² + k and drag k from −5 to 5: the parabola slides up and down. Enter y = (x + k)² and drag: the parabola slides left as k grows, right as k falls. Enter y = k · x² and drag k through 0: the parabola narrows, flattens to the horizontal axis at k = 0 and flips over as k turns negative. Enter y = |kx| and drag: the V narrows as |k| grows and reflects across the vertical axis when k changes sign, though the reflection is invisible for |x| because the V is symmetric already. A screenshot of two positions of the slider, labeled with the k values, is an explanation a reader can check.

Even and odd functions. A function is even if f(−x) = f(x) for every x in its domain; its graph is symmetric about the vertical axis, so the graph to the left of the axis is the mirror image of the graph to the right. A function is odd if f(−x) = −f(x) for every x; its graph is symmetric about the origin, so rotating the graph half a turn about the origin leaves it unchanged. From an algebraic expression, replace x by −x and simplify. For f(x) = x⁴ − 3x², f(−x) = (−x)⁴ − 3(−x)² = x⁴ − 3x² = f(x), so f is even; the check f(2) = 16 − 12 = 4 = f(−2) agrees. For g(x) = x³ − 5x, g(−x) = −x³ + 5x = −(x³ − 5x) = −g(x), so g is odd; g(2) = 8 − 10 = −2 and g(−2) = −8 + 10 = 2. For h(x) = x² + x, h(−x) = x² − x, which is neither h(x) nor −h(x), because h(2) = 6 and h(−2) = 2; h is neither even nor odd, and its graph has no symmetry about the vertical axis or the origin. A polynomial with only even powers of x (a constant counts as x⁰) is even, and one with only odd powers is odd, which is where the names come from; |x| is even without being a polynomial, and 1/x is odd. The misconception to avoid is that every function is one or the other: most functions, including every horizontally shifted parabola such as (x − 2)², are neither, and a horizontal shift f(x + k) with k ≠ 0 usually destroys the symmetry of an even function (a vertical shift f(x) + k does not).

Worked identification of effects: Starting from f(x) = x², describe the graph of each of x² + 5, (x + 1)², 3x² and (x − 4)² − 2. x² + 5 is f(x) + 5: shifted up 5, vertex (0, 5). (x + 1)² is f(x + 1): shifted left 1, vertex (−1, 0), through (0, 1) and (1, 4). 3x² is 3f(x): stretched vertically by 3, vertex (0, 0), through (1, 3) and (2, 12). (x − 4)² − 2 is f(x − 4) − 2, k = −4 in the horizontal replacement and k = −2 in the vertical: shifted right 4 and down 2, vertex (4, −2), through (5, −1) and (6, 2). Every value is checked by substitution: (6 − 4)² − 2 = 4 − 2 = 2.

Worked value of k from graphs: The graph of f(x) = x² and a second parabola with the same vertex are given. The second has vertex (0, 0) but passes through (2, −8) where f passes through (2, 4). The vertex did not move, so no shift occurred; the sign flipped and the height changed, so the replacement is kf(x). Divide: k = −8/4 = −2, and the check at x = 1 gives −2 · 1 = −2, so the second graph should pass through (1, −2). In a second picture the parabola has the same shape as f and its vertex is at (−3, −1). The horizontal move of 3 to the left is f(x + 3), so k = 3 there, and the vertical move of 1 down is a further + (−1): the function is (x + 3)² − 1, and the check (−2 + 3)² − 1 = 0 says it crosses the horizontal axis at x = −2 and, by symmetry, at x = −4.

Worked even and odd classification: Classify p(x) = 2x⁶ − x² + 7, q(x) = x⁵ + 4x³ and r(x) = x³ + 1. For p, replacing x by −x leaves every term unchanged because 6, 2 and 0 are even powers, so p(−x) = p(x) and p is even; its graph is symmetric about the vertical axis. For q, (−x)⁵ + 4(−x)³ = −x⁵ − 4x³ = −q(x), so q is odd and its graph is symmetric about the origin; q(1) = 5 and q(−1) = −5 confirm it. For r, r(−x) = −x³ + 1, and r(1) = 2 while r(−1) = 0: neither 2 nor −2, so r is neither even nor odd. The constant term 1 is an even power, x⁰, mixed with the odd power x³, which is why the symmetry fails.

Representation task: On one set of axes from x = −3 to x = 5 and y = 0 to y = 10, plot f(x) = x² using x = −3, −2, −1, 0, 1, 2, 3. Then plot f(x − 2) = (x − 2)² using x = −1 through 5 and 2f(x) = 2x² using x = −2 through 2. Label each curve with its formula and with the replacement rule that produced it (f(x + k) with k = −2, and kf(x) with k = 2). Beside the graph write the vertex of each parabola and one point on each curve, and explain in one sentence why (x − 2)² moved right although the formula shows a subtraction.

Independent performance: Algebra 1 and Math 2 route: use a quadratic such as (x − 1)² + 2 and produce the three screenshots or sketches f(x) + k, kf(x) and f(x + k) only, then complete the partner and even-or-odd parts below. Algebra 2 and Math 3 route: using a graphing tool, choose a function that is not centered at the origin, such as √(x − 1), |x − 2| or 2ˣ, so that f(kx) is not simply a vertical stretch, and produce four labeled screenshots or sketches showing f(x) + k, kf(x), f(kx) and f(x + k) for one positive and one negative value of k each. For every picture, state k, name the effect on the graph and verify it with one point whose coordinates you compute by hand. Then take two of your transformed graphs, hide the formulas, and give them to a partner to find k; check their answers against yours. Finally, classify four functions of your own, two polynomials and two others, as even, odd or neither, using both the algebraic test f(−x) and a description of the symmetry of the graph. Success criteria: the direction of every shift, stretch, compression and reflection matches the sign and size of k; each k found from a graph is confirmed by a second point; each even or odd classification shows the substitution of −x and names the symmetry.