6.2 Graphs of Functions, Vertical Line Test, Maxima and Minima

6.2 Graphs of Functions, Vertical Line Test, Maxima and Minima

Table 11’s Functions slice includes graphing functions. On AAF that means reading a graph as a function, rejecting a graph that is not a function, and naming maxima, minima, and range from a picture or a simple formula. Skills Insight band 276–300 adds connections among graphical, tabular, and algebraic representations of the absolute-value function.

The definition does not change when you move from a formula to a picture: a relation is a function when each input has exactly one output. On the coordinate plane the input is the x-coordinate. A vertical line holds x fixed and scans all possible y values. That is the vertical line test.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

The vertical line test

Draw or imagine a vertical line x = a sweeping across the graph.

• If every such line meets the graph in at most one point, the graph is a function.
• If any vertical line meets the graph in two or more points, the graph is not a function: that x-value would have two outputs.

“At most one” allows a vertical line to miss the graph entirely — those x-values are simply outside the domain. A function is allowed to skip inputs. It is not allowed to double them.

Which graphs are functions?

A circle fails because, for example, x = 0 meets (0, 3) and (0, −3). The equation x^2 + y^2 = 9 solved for y is y = ±√(9 − x^2), two outputs. Splitting into the upper semicircle y = √(9 − x^2) does make a function; the full circle does not.

AAF will not always print the equation. You may see four graphs and be asked which one represents a function. Hunt for any place the curve doubles back over the same x. Loops, circles, and left-opening parabolas are the usual failures. A capital-C shape fails; a capital-U shape passes.

Maxima and minima from graphs

A maximum value is the largest output; a minimum is the smallest. On a graph they are the highest and lowest points that exist on the piece of the graph you are given.

• A parabola opening up (a > 0 in y = ax^2 + bx + c) has a minimum at the vertex and no maximum on all real x.
• A parabola opening down (a < 0) has a maximum at the vertex and no minimum on all real x.
• A V-shaped absolute-value graph y = |x − h| + k has a minimum k at x = h if the V opens up, or a maximum k if the V is flipped (y = −|x − h| + k).
• On a closed interval, also check the endpoints; a linear piece on [1, 4] has its max and min at the ends.

Name the output as the max or min, then the x-location if asked. “The maximum is 5” is not the same sentence as “the maximum occurs at x = 2.” AAF tests both domain/range and max/min; mix them up and you lose the item even after you found the vertex.

Teach the vertex with a quadratic first

Let y = −x^2 + 4x + 1. This opens down, so it has a maximum. The vertex x-coordinate is x = −b/(2a) = −4 / (2(−1)) = 2. Then y = −(2)^2 + 4(2) + 1 = −4 + 8 + 1 = 5. The maximum value is 5, attained at x = 2. The range is y ≤ 5.

Now lift the idea to an even even-degree polynomial that opens down. Let y = −2x^4 + 7. For every real x, x^4 ≥ 0, so −2x^4 ≤ 0 and y = −2x^4 + 7 ≤ 7. Equality holds at x = 0. As |x| → ∞, y → −∞, so there is no minimum. The range is y ≤ 7. You do not need calculus on AAF: even degree, negative leading coefficient, and a constant term that is clearly the peak when the variable terms vanish is enough.

Compare y = 2x^4 + 7: now the range is y ≥ 7 (a global minimum of 7). The sign of the leading coefficient decides max versus min; the even degree decides that a global extreme exists at all. Odd-degree polynomials such as y = x^3 have neither a global max nor a global min on all real numbers, even though they are still functions (they pass the vertical line test).

From a table, scan the listed y values. If f is given by (−2, 1), (0, 7), (1, 5), (3, −4) and you are told these are the only critical samples on a closed interval that includes those x-values, the listed maximum is 7 and the listed minimum is −4. Do not assume a missing x-value hides a higher peak unless a formula is supplied.

Absolute value: V-shape, table, and formula

The parent graph y = |x| is a V with vertex at the origin, left ray slope −1, right ray slope 1. Shifts move the vertex: y = |x − h| + k has vertex (h, k). Example: y = |x + 3| − 2 is y = |x − (−3)| − 2, vertex (−3, −2). The graph never goes below −2, so the range is y ≥ −2.

Algebra, table, and graph must tell the same story — that is the 276–300 absolute-value connection.

• Algebra: y = |x − 1|.
• Table: at x = −2, −1, 0, 1, 2, 3 the outputs are 3, 2, 1, 0, 1, 2.
• Graph: V with vertex (1, 0), passing through those points.

If a table lists (1, 0) and (3, 2) but also (1, 4), the table cannot be this function — two outputs for x = 1, so the vertical line test already failed.

Piecewise, |x − 1| = 1 − x when x < 1 and x − 1 when x ≥ 1. Evaluating f(−4) for f(x) = |x − 1| is 5, not −5. Absolute value never returns a negative output; the expression inside the bars may be negative.

A reflected V, y = −|x − 1| + 4, has a maximum of 4 at x = 1 and range y ≤ 4. That is the same range shape as y = −2x^4 + 7, with a different graph. Matching a table to y = −|x − 1| + 4 means the peak 4 sits at x = 1 and the values fall by 1 for each integer step away from 1.

Reading max/min off a picture under time pressure

1. Decide whether the graph is a function (vertical line test) before you name a max.
2. Find the highest plotted y-value that the curve actually attains — that is the maximum value (an output), not the x-coordinate of the peak.
3. If arrows continue up, there is no maximum.
4. If arrows continue down, there is no minimum.
5. Report range in inequality form: y ≤ 7, y ≥ −2, or all real numbers.

Function-notation evaluation from Function Notation, Evaluation, Domain, and Range still applies on a graph: if the graph of f has a point (3, 8), then f(3) = 8 and, when f is one-to-one on that piece, f^{-1}(8) = 3.