Lesson 1.2 · Unit 1 · F-IF.7
The small family every other graph in the course is built from
Nearly every function in this course is a parent function that has been shifted, stretched or reflected. Knowing the parents by shape, domain, range and a few key points means a new graph is an exercise in recognition, not in plotting dozens of points.
- The constant and identity functions: \( y = c \) is a horizontal line, and \( y = x \) is the diagonal through the origin.
- The quadratic \( y = x^2 \) is a U-shape with vertex at the origin, domain all reals, range \( [0, \infty) \).
- The cubic \( y = x^3 \) has domain and range all reals, passes through \( (-1, -1) \), \( (0, 0) \) and \( (1, 1) \), and flattens at the origin.
- The absolute value \( y = |x| \) is a V with vertex at the origin, range \( [0, \infty) \).
- The square root \( y = \sqrt{x} \) starts at the origin and rises slowly, with domain and range both \( [0, \infty) \).
- The cube root \( y = \sqrt[3]{x} \) has domain and range all reals, because an odd root of a negative number exists.
- The reciprocal \( y = \dfrac{1}{x} \) has two branches and the axes as asymptotes; domain and range are all reals except 0.
- The exponential \( y = 2^x \) is always positive, passes through \( (0, 1) \), and has the \( x \)-axis as a horizontal asymptote.
Where students lose marks: confusing the range of \( y = x^2 \) with its domain. Every real number can be squared, but the outputs are never negative, so the domain is all reals and the range is only \( [0, \infty) \).
The problem. (a) State the domain and range of \( y = \sqrt{x} \) and of \( y = \dfrac{1}{x} \). (b) Give the coordinates of the points of \( y = x^3 \) at \( x = -2, -1, 0, 1, 2 \). (c) Which parents are increasing on the whole real line? (d) Compare the growth of \( y = x^2 \) and \( y = 2^x \) for \( x = 1 \) to 5.
Step one: the square root for (a). The radicand must be nonnegative, so the domain is \( [0, \infty) \). A square root is never negative, so the range is also \( [0, \infty) \).
Step two: the reciprocal. Division by zero is undefined, so the domain is \( x \ne 0 \). Can the output be zero? \( \dfrac{1}{x} = 0 \) has no solution, since a fraction with numerator 1 is never zero. So the range is \( y \ne 0 \). Domain and range are both \( (-\infty, 0) \cup (0, \infty) \).
Step three: the cubic points for (b). \( (-2)^3 = -8 \), \( (-1)^3 = -1 \), \( 0 \), \( 1 \), \( 2^3 = 8 \). Points: \( (-2, -8) \), \( (-1, -1) \), \( (0, 0) \), \( (1, 1) \), \( (2, 8) \). Notice the symmetry: the point at \( -2 \) is the negative of the point at 2.
Step four: read the shape from the points. From \( -1 \) to 0 the output rises by 1, but from 1 to 2 it rises by 7. The graph is steep at the ends and flat near the origin, which is the cubic's characteristic shape.
Step five: increasing functions for (c). A function is increasing on the whole line if a larger input always gives a larger output. That rules out \( x^2 \) and \( |x| \), which fall before they rise, and \( 1/x \), which is not defined at 0. The increasing parents are \( y = x \), \( y = x^3 \), \( y = \sqrt[3]{x} \) and \( y = 2^x \). The constant is neither increasing nor decreasing.
Step six: tabulate for (d). At \( x = 1 \): \( 1 \) against \( 2 \). At \( x = 2 \): \( 4 \) against \( 4 \). At \( x = 3 \): \( 9 \) against \( 8 \). At \( x = 4 \): \( 16 \) against \( 16 \). At \( x = 5 \): \( 25 \) against \( 32 \).
Step seven: interpret the crossings. The two graphs meet at \( x = 2 \) and \( x = 4 \), and \( 2^x \) is larger at \( x = 5 \). Between 2 and 4 the quadratic is ahead. There is also a third crossing for a negative \( x \), near \( -0.767 \), where both equal about 0.588.
Step eight: state what lasts. After \( x = 4 \), \( 2^x \) stays ahead forever. Doubling beats any fixed power in the long run, which is the first instance of a principle this course uses repeatedly: to compare functions, look at their behavior at the ends. A table of small values can mislead, since here the quadratic led at \( x = 3 \) and lost by \( x = 5 \).
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- State the domain of \( y = \sqrt{x} \).
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The radicand cannot be negative. \( [0, \infty) \)
- State the range of \( y = |x| \).
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An absolute value is never negative, and every nonnegative number is reached. \( [0, \infty) \)
- Name the asymptotes of \( y = \dfrac{1}{x} \).
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The function blows up as \( x \to 0 \) and levels off as \( |x| \) grows. \( x = 0 \) (vertical) and \( y = 0 \) (horizontal)
- Evaluate \( 2^x \) at \( x = -1 \).
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A negative exponent means a reciprocal: \( 2^{-1} = \dfrac{1}{2} \). \( \dfrac{1}{2} \)
- Which parent has domain and range all real numbers, is not a line, and has a flat spot at
the origin?
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\( y = x^2 \) has a restricted range and \( y = x \) is a line. The cubic is defined for all inputs, takes all outputs, and flattens at the origin. \( y = x^3 \)
- Which of \( x^2 \), \( x^3 \), \( |x| \), \( 2^x \) are increasing on the whole real line?
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\( x^2 \) and \( |x| \) decrease for negative \( x \), so they are out. \( x^3 \) and \( 2^x \)
- Solve \( x^2 = 2^x \) for positive \( x \) and verify.
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Try small integers: \( x = 2 \) gives \( 4 = 4 \), and \( x = 4 \) gives \( 16 = 16 \). Both check. No other positive integer works: \( x = 3 \) gives \( 9 \ne 8 \), and \( x = 5 \) gives \( 25 \ne 32 \). \( x = 2 \) and \( x = 4 \)
- For \( 0 \lt x \lt 1 \), which is larger, \( \sqrt{x} \) or \( x \)? Test and explain.
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Take \( x = 0.25 \): \( \sqrt{0.25} = 0.5 \), which exceeds 0.25. Take \( x = 0.81 \): \( \sqrt{0.81} = 0.9 \gt 0.81 \). Squaring a number between 0 and 1 makes it smaller, so taking the root makes it larger. For \( x \gt 1 \) the order reverses: \( \sqrt{4} = 2 \lt 4 \). \( \sqrt{x} \gt x \) on \( (0, 1) \)
- Give the points of \( y = \sqrt[3]{x} \) at \( x = -8, -1, 0, 1, 8 \).
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\( \sqrt[3]{-8} = -2 \), \( \sqrt[3]{-1} = -1 \), \( 0 \), \( 1 \), \( \sqrt[3]{8} = 2 \). \( (-8, -2), (-1, -1), (0, 0), (1, 1), (8, 2) \)
- Why does \( \sqrt{x} \) have a restricted domain but \( \sqrt[3]{x} \) does not?
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A square root asks for a number whose square is \( x \). Squares of real numbers are never negative, so no real number answers for a negative \( x \). A cube root asks for a number whose cube is \( x \). Cubes can be negative, since \( (-2)^3 = -8 \), so every real \( x \) has exactly one real cube root. The rule generalizes: even roots need a nonnegative radicand, odd roots do not. Even powers cannot be negative; odd powers can