Math by First Principles
Chapter Review · a rapid pre-test review (speedrun)

Graphing Functions: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Graph of a function
The set of all points (x,f(x))(x, f(x)): each point's first coordinate is an input, its second that input's output.
Zeros (roots) of a function
The inputs whose output is 00, that is, the first coordinates of the xx-intercepts.
Turning point
Where the graph stops falling and starts rising, or the reverse: a minimum at the bottom of a valley, a maximum at the top of a hill.
Shift (translation)
A move of every point the same distance in the same direction. The shape is untouched.
Stretch and compression
A rescaling of every height, or of every horizontal position, by one common factor: a stretch pulls the graph away from an axis, a compression presses it in.
Reflection
A flip across a line. Points on that line are the hinge and do not move.
Fixed point
A point a transformation leaves where it was, such as a point of the line y=xy = x under reflection across it.

Formulas and theorems

  • Reading a graph in both directions

    (a,b) is on the graph of f  ⟺  f(a)=b\begin{gathered} (a, b) \text{ is on the graph of } f \\ \iff f(a) = b \end{gathered}

    Use when Needs aa in the domain. Read backwards, the solutions of f(x)=bf(x) = b are the inputs directly at the meeting points of the line y=by = b: several, one, or none.

  • Intercepts and zeros

    If 00 is in the domain, the yy-intercept is (0,f(0))(0, f(0)); otherwise there is no yy-intercept. An xx-intercept is a point (a,0)(a, 0) on the graph, so the xx-intercepts come from solving f(x)=0f(x) = 0.

    Use when At most ONE yy-intercept; xx-intercepts may be several, one, or none. Zero the coordinate of the axis met, the opposite letter.

    e.g. f(x)=x2−x−6f(x) = x^2 - x - 6: yy-intercept (0,−6)(0, -6), zeros x=3x = 3 and x=−2x = -2.

  • Domain and range from a graph

    The domain is the graph's horizontal spread along the xx-axis; the range is its vertical spread along the yy-axis.

    Use when Arrowheads mean that spread runs on forever; a graph drawn between two endpoints stops at them.

  • Increasing and decreasing

    increasing: a<b  ⟹  f(a)<f(b)decreasing: a<b  ⟹  f(a)>f(b)\begin{gathered} \text{increasing:}\ a < b \implies f(a) < f(b) \\ \text{decreasing:}\ a < b \implies f(a) > f(b) \end{gathered}

    Use when Stated on an interval and always read left to right, in the direction of increasing xx: the graph rises where ff increases and falls where it decreases.

  • Shifts

    y=f(x)+k  moves up by ky=f(x−h)  moves right by h\begin{gathered} y = f(x) + k \ \text{ moves up by } k \\ y = f(x - h) \ \text{ moves right by } h \end{gathered}
    A shift moves every point right h and up k, and leaves the shape aloneTwo identical U-shaped curves against faint axes, the second placed up and to the right of the first. A dot marks the lowest point of each. From the left dot a highlighted horizontal segment labeled h runs to the right, then a highlighted vertical segment labeled k runs up to the right dot, so the move splits into a horizontal part h and a vertical part k while the shape stays the same.hk(a, b)(a + h, b + k)
    Text description

    A curve and its translate, with a horizontal leg h and a vertical leg k carrying the point (a, b) to the point (a + h, b + k).

    Use when INSIDE (x−hx - h) is HORIZONTAL and runs OPPOSITE to the sign shown: (a,b)→(a+h,b)(a, b) \to (a + h, b), moving the domain and the xx-intercepts. OUTSIDE (+k+ k) is VERTICAL and runs WITH the sign: (a,b)→(a,b+k)(a, b) \to (a, b + k), moving the range. A vertical shift can change HOW MANY xx-intercepts there are.

  • Stretches and compressions

    y=a f(x)  sends  (x0,y0)→(x0, ay0)y=f(bx)  sends  (x0,y0)→(x0b, y0)\begin{gathered} y = a\,f(x) \ \text{ sends } \ (x_0, y_0) \to (x_0,\ a y_0) \\ y = f(bx) \ \text{ sends } \ (x_0, y_0) \to \left(\tfrac{x_0}{b},\ y_0\right) \end{gathered}

    Use when a≠0a \neq 0 and b≠0b \neq 0, and a negative factor reflects as well as scales. OUTSIDE, aa scales heights DIRECTLY: taller when ∣a∣>1\lvert a \rvert > 1, shorter when 0<∣a∣<10 < \lvert a \rvert < 1. INSIDE, bb scales widths by the RECIPROCAL 1b\tfrac{1}{b}, so ∣b∣>1\lvert b \rvert > 1 COMPRESSES the graph toward the yy-axis while 0<∣b∣<10 < \lvert b \rvert < 1 STRETCHES it away. Vertical scaling reshapes the range and pins every xx-intercept; horizontal scaling reshapes the domain and pins the yy-intercept.

    e.g. (4,6)(4, 6) becomes (4,−3)(4, -3) on y=−12f(x)y = -\tfrac{1}{2}f(x).

  • The two reflections on their own

    y=−f(x)  flips across the x-axisy=f(−x)  flips across the y-axis\begin{gathered} y = -f(x) \ \text{ flips across the } x\text{-axis} \\ y = f(-x) \ \text{ flips across the } y\text{-axis} \end{gathered}
    The outside minus flips across the x-axis, the inside minus across the y-axisA short rising curve sits in the upper right quarter of a pair of axes and is labeled y equals f of x. Directly beneath it, across the horizontal axis, a highlighted copy is flipped top to bottom and labeled y equals minus f of x. Directly opposite it, across the vertical axis, a second highlighted copy is flipped left to right and labeled y equals f of minus x. All three curves have the same shape.y = f(x)y = −f(x)y = f(−x)xy
    Text description

    One curve with two highlighted mirror images of it: the flip across the x-axis, which is the graph of minus f of x, and the flip across the y-axis, which is the graph of f of minus x.

    Use when The cases a=−1a = -1 and b=−1b = -1: the outside minus negates the output, the inside minus the input. Neither changes the shape.

    e.g. (5,−2)(5, -2) becomes (5,2)(5, 2) on y=−f(x)y = -f(x) and (−5,−2)(-5, -2) on y=f(−x)y = f(-x).

  • Scaling and shifting together, and vertex form

    y=a f(x−h)+ksends(p,q)→(p+h, aq+k)\begin{gathered} y = a\,f(x - h) + k \\ \text{sends} \quad (p, q) \to (p + h,\ aq + k) \end{gathered}

    Use when a≠0a \neq 0; taking a=1a = 1 leaves the pure combined shift, right hh and up kk. On the output the ORDER is scale first, then add: aq+kaq + k, never a(q+k)a(q + k). For a parabola this is vertex form y=a(x−h)2+ky = a(x - h)^2 + k, vertex (h,k)(h, k), aa setting width and opening.

    e.g. (4,5)(4, 5) becomes (5,13)(5, 13) on y=3f(x−1)−2y = 3f(x - 1) - 2.

  • Graph of an inverse

    (a,b) on the graph of f  ⟺  (b,a) on the graph of f−1\begin{gathered} (a, b) \text{ on the graph of } f \\ \iff (b, a) \text{ on the graph of } f^{-1} \end{gathered}
    The graph of the inverse is the graph of f mirrored in the line y = xA rising curve labeled f lies above a highlighted dashed diagonal labeled y equals x, and a second rising curve labeled f inverse lies below it, the same shape reflected. A dot on the upper curve labeled (a, b) and a dot on the lower curve labeled (b, a) are joined by a short highlighted dashed segment that meets the diagonal at a right angle and is cut in half by it.(a, b)(b, a)ff−1y = x
    Text description

    The graph of f and the graph of its inverse as mirror images in the dashed line y equals x, with the point (a, b) on one carried to the point (b, a) on the other.

    Use when Swapping a point's coordinates reflects it across y=xy = x, so the graph of f−1f^{-1} is the graph of ff mirrored in that diagonal, no formula needed. That mirror image is the graph of a FUNCTION exactly when ff is one-to-one, which is what f−1f^{-1} needs in order to exist at all, and f(c)=cf(c) = c forces f−1(c)=cf^{-1}(c) = c.

    e.g. (9,−2)(9, -2) on ff gives (−2,9)(-2, 9) on f−1f^{-1}.

  • Domain and range swap under inversion

    The domain of f−1f^{-1} is the range of ff, and the range of f−1f^{-1} is the domain of ff.

    Use when Whenever f−1f^{-1} exists: reflecting across y=xy = x trades the two axes, so a horizontal spread becomes a vertical one. Never carry an interval across unchanged.

    e.g. Domain 3≤x≤73 \le x \le 7, range 0≤y≤40 \le y \le 4 inverts to domain 0≤x≤40 \le x \le 4, range 3≤y≤73 \le y \le 7.

Problem types, step by step

Graph a function from a table of values

  1. Choose several allowed inputs spread across the domain: a negative one, 00, and a few positives when the domain permits them.
  2. Evaluate the rule at each, wrapping the input in parentheses so signs survive.
  3. Plot the pairs and join them only when the rule's type is already known: a ruler for a known linear rule, a smooth parabola for a known quadratic rule, and otherwise leave the points unconnected.
  4. Recheck any point that misses the shape the others make.

e.g. f(x)=x2−4f(x) = x^2 - 4 gives (−1,−3)(-1, -3), (0,−4)(0, -4), (2,0)(2, 0).

Read values, intercepts, domain, and range off a graph

  1. For f(a)f(a), start at aa on the horizontal axis, move to the curve, and read the height.
  2. For f(x)=bf(x) = b, start at bb on the vertical axis instead, slide across the line y=by = b, and read the input beneath EVERY meeting point.
  3. Take the points where the graph meets the two axes for the intercepts, or evaluate f(0)f(0) when 00 is in the domain and solve f(x)=0f(x) = 0.
  4. Sweep the horizontal spread for the domain and the vertical spread for the range, noting endpoints against arrowheads.

Find where a function increases and decreases

  1. Trace the graph left to right, in the direction of increasing xx.
  2. Mark every turning point, and call it a maximum or a minimum by whether the curve peaks or bottoms out.
  3. Report the stretches of INPUTS, read off the horizontal axis, where it rises and where it falls.

Describe the transformation a rule performs

  1. Split the rule into what sits OUTSIDE the function and what sits INSIDE.
  2. Read the outside as a vertical scaling then a vertical shift, running with intuition.
  3. Read the inside as a horizontal scaling then a horizontal shift, running opposite to it.
  4. Note any negative factor as a reflection as well.

e.g. y=−2f(x+1)y = -2f(x + 1) is a shift left 11, a vertical stretch by 22, and a flip across the xx-axis.

Move a point, or a whole graph, through a transformation

  1. Send the first coordinate through the inside change and the second through the outside change.
  2. If the inside is more than a plain xx, set the WHOLE inside expression equal to the old input and solve.
  3. For a whole graph, run a few anchors (turning point, intercepts) through the same map and redraw the shape.
  4. Confirm the pinned points held: a vertical scaling fixes every xx-intercept, a horizontal scaling the yy-intercept.

e.g. (6,2)(6, 2) lands at (3,−4)(3, -4) on y=−2f(2x)y = -2f(2x).

Track a domain and range through a transformation

  1. Decide whether the change sits inside or outside the function.
  2. Apply an inside change to the domain endpoints only, an outside change to the range endpoints only.
  3. Rewrite the interval from its smaller value up, since a negative factor swaps the endpoints.

e.g. Range −2≤y≤6-2 \le y \le 6 becomes −12≤y≤4-12 \le y \le 4 under y=−2f(x)y = -2f(x).

Sketch the graph of f−1f^{-1} from the graph of ff

  1. Confirm ff passes the horizontal line test, so an inverse function exists at all.
  2. Draw y=xy = x with equal scales on the axes, or the diagonal is not a true mirror.
  3. Read off several points of ff and swap BOTH coordinates of each.
  4. Mark any crossing of y=xy = x as fixed, then draw the mirrored curve.

e.g. (−2,−3)(-2, -3), (0,1)(0, 1), (3,4)(3, 4) on gg give (−3,−2)(-3, -2), (1,0)(1, 0), (4,3)(4, 3) on g−1g^{-1}.

Graph the inverse of a function that fails the horizontal line test

  1. Find a branch on which the rule is one-to-one and restrict the domain to it.
  2. Reflect ONLY that branch across y=xy = x.
  3. State the restriction with the answer, since the branch's outputs are the inverse's allowed inputs.

e.g. y=x2y = x^2 on x≥0x \ge 0 reflects to y=xy = \sqrt{x}, carrying (3,9)(3, 9) to (9,3)(9, 3).

Exam traps

  • Trap Sending y=f(x−4)y = f(x - 4) LEFT because of the minus sign, so the vertex of y=(x−4)2y = (x - 4)^2 gets plotted at (−4,0)(-4, 0).

    Fix An inside change runs OPPOSITE to its sign: this moves RIGHT 44, putting that vertex at (4,0)(4, 0). Read hh as the value making the inside zero, so y=(x+3)2−5y = (x + 3)^2 - 5 has vertex (−3,−5)(-3, -5).

  • Trap Reading y=f(3x)y = f(3x) as a stretch to three times the width.

    Fix An inside factor acts by its RECIPROCAL. Inputs are divided by 33, compressing the graph toward the yy-axis, so a zero at x=12x = 12 lands at x=4x = 4. Only y=f ⁣(x3)y = f\!\left(\tfrac{x}{3}\right) widens it.

  • Trap Letting a vertical stretch drag the xx-intercepts inward with it, because the stretched graph does look narrower.

    Fix Only heights change, and a×0=0a \times 0 = 0, so every zero is pinned. Stretching f(x)=x2−4f(x) = x^2 - 4 into y=3x2−12y = 3x^2 - 12 keeps both crossings at x=−2x = -2 and x=2x = 2; what moved is the yy-intercept, from (0,−4)(0, -4) to (0,−12)(0, -12). A horizontal scaling does the reverse, pinning the yy-intercept and sliding the zeros.

  • Trap Reflecting across an axis, or negating just one coordinate, to build an inverse.

    Fix The mirror for an inverse is the diagonal y=xy = x, and BOTH coordinates trade: (a,b)→(b,a)(a, b) \to (b, a). Reflecting across the xx-axis produces y=−f(x)y = -f(x), a different graph entirely.

  • Trap Reflecting the whole graph of a function that fails the horizontal line test and calling the result f−1f^{-1}.

    Fix That reflection fails the vertical line test, so it is not a function. Restrict to a one-to-one branch first: all of y=x2y = x^2 reflects to a sideways curve the line x=4x = 4 meets twice, at (4,2)(4, 2) and (4,−2)(4, -2).

Chapter Test Questions from across the chapter