Graph linear, quadratic and simple exponential functions from their formulas and indicate intercepts, maxima, minima and end behavior

Learning goal: Graph functions expressed symbolically and indicate the key features of the graph: for linear functions the intercepts; for quadratic functions the intercepts, maximum or minimum; for simple exponential functions the intercept and end behavior. Do this by hand in simple cases and with graphing technology for more complicated cases, and connect each key feature to a characteristic of the situation the function models so that the choice of a linear, quadratic or exponential model is justified.

Graphing is finding the features first. A graph drawn by plotting random points is slow and unreliable. A graph drawn from its key features is fast and self-checking: locate the intercepts, locate any turning point, decide the end behavior, and then a few extra points confirm the shape. Each family of functions has its own list of features to find, and the list is short.

Linear functions. A linear function f(x) = mx + b has a straight-line graph with slope m and vertical intercept (0, b). Its horizontal intercept is where mx + b = 0, at x = −b/m, provided m ≠ 0. Take f(x) = 2x − 6. The vertical intercept is (0, −6). The horizontal intercept satisfies 2x − 6 = 0, so x = 3, giving (3, 0). Two points determine the line; a third, f(6) = 6, checks it. The slope 2 means the output rises 2 for each unit increase in input, and the end behavior is that f(x) increases without bound as x increases and decreases without bound as x decreases. A horizontal line f(x) = b has slope 0, a vertical intercept at b and no horizontal intercept unless b = 0. In a situation, the vertical intercept is the starting value and the slope is the constant rate, and a linear model is the appropriate selection when the quantity changes by the same amount each step.

Quadratic functions. A quadratic function g(x) = ax² + bx + c, with a ≠ 0, has a parabola for its graph. If a > 0 the parabola opens upward and its vertex is a minimum; if a < 0 it opens downward and its vertex is a maximum. The vertical intercept is (0, c). The horizontal intercepts, if any, are the solutions of ax² + bx + c = 0, found by factoring when factoring is reasonable and by the quadratic formula otherwise. The vertex lies on the axis of symmetry x = −b/(2a), which is also halfway between the two horizontal intercepts when they exist.

A fully worked model: graph g(x) = x² − 4x − 5. Here a = 1 > 0, so the parabola opens upward and has a minimum. Vertical intercept: g(0) = −5, so (0, −5). Horizontal intercepts: x² − 4x − 5 = (x − 5)(x + 1) = 0, so x = 5 or x = −1, giving (5, 0) and (−1, 0). Axis of symmetry: x = −(−4)/(2 · 1) = 2, which is indeed halfway between −1 and 5. Vertex: g(2) = 4 − 8 − 5 = −9, so the minimum is −9 at x = 2. Two more points by symmetry: g(1) = g(3) = −8 and g(−2) = g(6) = 7. End behavior: the x² term dominates, so g(x) increases without bound as x goes to either extreme. Plot the seven points and draw the smooth U shape through them. If the situation is a downward parabola, such as the fenced area A(x) = −2x² + 8x, the same steps give intercepts x = 0 and x = 4 from −2x(x − 4) = 0, an axis at x = 2, and a maximum A(2) = −8 + 16 = 8; the maximum is the largest possible area and the intercepts are the dimensions that give no area at all.

When there are no horizontal intercepts. The parabola y = x² + 1 has vertex (0, 1), a minimum of 1, and opens upward, so it never reaches the horizontal axis; x² + 1 = 0 has no real solution. A graph with no horizontal intercept is not a mistake. Report it: the function is positive for every input.

Simple exponential functions. An exponential function h(x) = a · bˣ with a > 0 and b > 0, b ≠ 1, has vertical intercept (0, a), because b⁰ = 1, and no horizontal intercept, because a positive number times any power of a positive base is positive, never zero. If b > 1 the function is increasing: as x increases the outputs grow without bound, and as x decreases the outputs approach 0, so the horizontal axis is a horizontal asymptote on the left. If 0 < b < 1 the function is decreasing, growing without bound to the left and approaching the asymptote y = 0 on the right. For h(x) = 3 · 2ˣ: h(−2) = 0.75, h(−1) = 1.5, h(0) = 3, h(1) = 6, h(2) = 12, h(3) = 24. Each step right doubles the output, and the graph hugs the axis on the left and climbs steeply on the right. For k(x) = 16 · (1/2)ˣ: k(0) = 16, k(1) = 8, k(2) = 4, k(3) = 2, k(4) = 1, halving each step and approaching but never reaching 0. In a situation, a is the starting amount and b is the growth or decay factor per unit of input, and an exponential model is the right selection when the quantity is multiplied by the same factor each step.

By hand or with technology. The functions above are simple cases, meant to be graphed by hand from their features. When the coefficients are awkward, such as g(x) = 1.7x² − 3.9x + 0.4, or when the question is where two graphs cross, graphing technology draws the curve and locates intercepts and the vertex numerically. Technology does not remove the need to know the features: you still have to know that a parabola has one vertex, that an exponential function has an asymptote, and that a window showing only x from 0 to 1 might hide the horizontal intercept at x = 3. Reading a technology graph is an act of checking it against what the formula predicts.

Selecting the model from the features. The connection runs both ways. Given a situation, its characteristics choose the family: a constant rate means linear, a single maximum or minimum with symmetric behavior on either side suggests quadratic, a constant percent change means exponential. Given a graph, and the three candidate families of this lesson, its features suggest the family: a straight line, a single turning point with mirror symmetry, or a curve with a horizontal asymptote on one side and unbounded growth on the other. These features support a choice rather than prove it: y = x⁴ and y = |x| also have a single symmetric turning point, so check the shape and, from a table, the second differences. The mistake to avoid is treating a steeply rising curve as exponential simply because it is steep; the parabola y = x² is also steep for large x, and the test is the asymptote and the constant ratio, not the steepness.

Worked linear graph: Graph f(x) = −3x + 9. Vertical intercept: (0, 9). Horizontal intercept: −3x + 9 = 0 gives x = 3, so (3, 0). A check point: f(5) = −15 + 9 = −6. The line falls 3 units for each unit to the right, passes through (0, 9), (3, 0) and (5, −6), and its end behavior is falling without bound to the right and rising without bound to the left. If f models the water in a tank in liters after x minutes of draining, 9 liters is the starting amount, 3 liters per minute is the drain rate and (3, 0) is the moment the tank is empty, after which the model stops.

Worked quadratic graph: Graph q(x) = −x² + 6x − 1. Since a = −1 < 0, the parabola opens downward with a maximum. Vertical intercept (0, −1). Axis of symmetry x = −6/(2 · (−1)) = 3, and q(3) = −9 + 18 − 1 = 8, so the maximum is 8 at the vertex (3, 8). The horizontal intercepts solve −x² + 6x − 1 = 0; factoring is not reasonable, so the quadratic formula gives x = (−6 ± √(36 − 4))/(−2) = 3 ∓ √8 ≈ 0.17 and 5.83. Symmetric points: q(2) = q(4) = 7, q(1) = q(5) = 4, q(0) = q(6) = −1. The graph rises from the lower left to (3, 8) and falls to the lower right, and the end behavior is downward on both sides.

Worked exponential graph: Graph k(x) = 16 · (1/2)ˣ. Vertical intercept (0, 16). Table: k(1) = 8, k(2) = 4, k(3) = 2, k(4) = 1, k(−1) = 32. There is no horizontal intercept, because 16 · (1/2)ˣ is positive for every x. As x increases the graph drops toward the horizontal asymptote y = 0, halving with each step; as x decreases it climbs without bound, doubling with each step left. If k models the amount of a substance in grams after x days, it starts at 16 grams and halves every day, a decay model chosen precisely because the daily change is a constant factor rather than a constant amount.

Representation task: On one set of axes covering −2 ≤ x ≤ 6 and −10 ≤ y ≤ 10, graph f(x) = 2x − 6 and g(x) = x² − 4x − 5. For each function, mark and label every intercept and, for g, the vertex, and write the end behavior beside each graph. Read from the graph the inputs where the two functions appear to have the same output, then confirm by solving x² − 4x − 5 = 2x − 6, which simplifies to x² − 6x + 1 = 0 and has solutions x = 3 ± √8 ≈ 0.17 and 5.83.

Independent performance: Write one linear, one quadratic and one exponential function, each modeling a situation you describe in a sentence. Graph all three by hand, indicating intercepts, any maximum or minimum, and end behavior, and then check each graph with graphing technology. For each function write which feature of the situation made that family the appropriate choice. Success criteria: every intercept is computed algebraically and matches the drawing; the vertex of the quadratic is found from the axis of symmetry and verified by symmetric points; the exponential graph shows its asymptote and has no horizontal intercept; end behavior is stated in words for all three; and the justification for the model type names a characteristic of the situation, such as constant rate, single maximum or constant factor.