2.4 Key features of graphs

Intercepts

The intercepts are where a graph meets the axes. They are usually the easiest features to find and the first to mark on a sketch.

• The $y$-intercept is found by substituting $x = 0$. The graph of a function has at most one $y$-intercept, since $x = 0$ has at most one output.

• An x-intercept is found by substituting $y = 0$ and solving $f(x) = 0$. A graph may have no $x$-intercepts, one, or many.

At an $x$-intercept, a graph may cross the axis or just touch it and turn back. For example, $y = (x - 1)^2$ touches the $x$-axis at $(1, 0)$, while $y = x - 1$ crosses it there.

Zeros of functions and roots of equations

The $x$-intercepts of a graph connect two ideas that are often confused: zeros, which belong to functions, and roots, which belong to equations.

Any equation can be rewritten with zero on one side, so solving an equation is the same as finding the zeros of a function. The equation $x^3 = 2x + 5$ has the same roots as the zeros of $f(x) = x^3 - 2x - 5$. This is how a GDC solves equations that cannot be solved by algebra.

Maximum and minimum points

A graph that rises and then falls has a peak, and one that falls and then rises has a dip. These turning points are among the most useful features of a graph, because they give the largest and smallest values in a region: the most profit, the highest temperature, the least cost.

The word "local" means that the point is highest or lowest only compared with its neighbourhood. A local maximum can be lower than another part of the graph far away, and a local minimum can be higher than another part. The greatest and least values over the whole domain are described separately.

On the domain $-2 \le x \le 2.5$, the function $f(x) = x^3 - 3x$ has a local maximum point at $(-1, 2)$ and a local minimum point at $(1, -2)$. The global maximum is $8.125$, at the endpoint $x = 2.5$, which is higher than the local maximum. The global minimum is $-2$, reached both at the local minimum $x = 1$ and at the endpoint $x = -2$. On the domain $\mathbb{R}$, the same function has no global maximum or minimum at all, since $x^3 - 3x$ grows without bound in both directions.

Maximum value and maximum point

A maximum point is given by its coordinates; the maximum value is its $y$-coordinate, the largest output. For $f(x) = x^3 - 3x$, the local maximum point is $(-1, 2)$ and the local maximum value is $2$, which occurs when $x = -1$. The same distinction applies to minimum points and values.

Finding maximum and minimum points with technology

On a GDC, use the maximum or minimum function and choose a region of the graph around the turning point. The calculator returns both coordinates. Record them to 3 significant figures unless told otherwise, and check that the window shows every turning point in the domain.

The range of a continuous function on an interval runs from its global minimum to its global maximum, which is why turning points and endpoints are the features to check when a range is asked for (2.2). Finding turning points exactly, without technology, is done with calculus in 5.8.

Symmetry

A graph has line symmetry about a vertical line $x = a$ if the part of the graph on one side of the line is the mirror image of the part on the other side. In terms of the function, points equally far either side of $x = a$ have the same height:

Every quadratic graph, a parabola, has a vertical axis of symmetry through its turning point, called the vertex (2.6). Since the two zeros of a quadratic are mirror images in this axis, the axis lies halfway between them.

Some graphs have a different kind of symmetry. A graph symmetric in the $y$-axis satisfies $f(-x) = f(x)$, as for $y = x^2$ and $y = x^4 - 3x^2$. A graph with rotational symmetry of order 2 about the origin (it looks the same after a half-turn) satisfies $f(-x) = -f(x)$, as for $y = x^3$ and $y = x^3 - 3x$. These two types are studied as even and odd functions in 2.14 (HL).

Asymptotes

Some graphs approach a straight line more and more closely without reaching it, as the graph heads off towards infinity. Such a line is an asymptote.

The arrow notation is read "tends to". The statement "as $x \to \infty$, $f(x) \to 2$" means that when $x$ becomes very large, $f(x)$ gets as close to $2$ as desired.

Vertical asymptotes

A vertical asymptote often occurs at a value of $x$ that makes a denominator zero. Near $x = a$, the function divides by a number close to zero, so its values become very large in size. The function is not defined at $x = a$ itself, so its graph never crosses a vertical asymptote.

Horizontal asymptotes

A horizontal asymptote describes the long-run behaviour of a function. A table of values for large positive and large negative $x$ shows it: for $f(x) = \frac{2x + 1}{x - 1}$, $f(100) = \frac{201}{99} = 2.03\ldots$ and $f(1000) = \frac{2001}{999} = 2.003\ldots$, approaching $2$. Unlike a vertical asymptote, a horizontal asymptote may be crossed at finite values of $x$; it only describes what happens far to the left or right.

Exponential graphs also have horizontal asymptotes: $y = 2^x$ approaches $y = 0$ as $x \to -\infty$, as the graph of $g(x) = 2^x - 4$ approaches $y = -4$ (2.3, 2.9). Asymptotes of rational functions are found algebraically in 2.8.

Points of intersection

A point where two graphs meet has coordinates that satisfy both equations. Its $x$-coordinate is a solution of the equation $f(x) = g(x)$, and its $y$-coordinate is the common value. For two lines, or a line and a simple curve, the intersection can be found algebraically (2.1, 2.7); in general a GDC is used.

There are two methods with technology.

1. Intersect: graph $y = f(x)$ and $y = g(x)$ and use the intersection function at each crossing. This gives both coordinates directly.

2. Zero of the difference: graph $y = f(x) - g(x)$ and find its zeros. These are the $x$-coordinates of the intersections; substitute back to find each $y$-coordinate.

Key features and the number of solutions

The number of solutions of an equation $f(x) = k$ is the number of times the horizontal line $y = k$ meets the graph of $y = f(x)$. Key features decide this number. Lines through a maximum or minimum point touch the graph there, and the number of solutions changes as $k$ passes through a turning value.

Summary of key features

Exam-style questions on key features

A modelling question asks for the same features, but each answer must be interpreted in the context, with units and a sensible degree of accuracy.

Self-check

• To do:

  I can find the $x$- and $y$-intercepts of a graph by hand and with a GDC.

• To do:

  I can explain the difference between a zero of a function, a root of an equation and an $x$-intercept.

• To do:

  I can find local maximum and minimum points with a GDC and give their coordinates to 3 significant figures.

• To do:

  I can find the global maximum and minimum of a function on a domain, checking the endpoints.

• To do:

  I can use a line of symmetry to find mirror-image points and the vertex.

• To do:

  I can identify vertical and horizontal asymptotes and write their equations.

• To do:

  I can find all the points of intersection of two graphs with technology without missing any.

• To do:

  I can use key features to state a range and to count the solutions of $f(x) = k$.

1. (a)

   Show that $q = \cos\theta + \sqrt{4 - \sin^2\theta}$.

   [3]3 marks

   Markscheme (a)

   $\mathrm{PQ}^2 = (q - \cos\theta)^2 + \sin^2\theta$

2. (b)

   Find the exact value of $q$ when $\theta = \dfrac{5\pi}{6}$.

   [2]2 marks

   Markscheme (b)

   $q = -\dfrac{\sqrt{3}}{2} + \sqrt{4 - \dfrac{1}{4}}$

3. (c)

   Write down the maximum value of $q$ and the minimum value of $q$.

   [2]2 marks

   Markscheme (c)

   maximum $q = 3$ A1

   minimum $q = 1$

4. (d)

   Find the values of $\theta$ for which $q = 2.5$.

   [3]3 marks

   Markscheme (d)

   EITHER

   $\cos\theta + \sqrt{4 - \sin^2\theta} = 2.5$

5. (e)

   For $0 < \theta < \pi$, show that $\sin\left(\mathrm{O\hat{Q}P}\right) = \dfrac{1}{2}\sin\theta$

6. (f)

   Hence, write down the maximum size of $\mathrm{O\hat{Q}P}$ and the value of $\theta$ at which it occurs.

   [2]2 marks

   Markscheme (f)

   $\dfrac{\pi}{6}$

The equation $x^3 = 2x + 5$ rearranges to $x^3 - 2x - 5 = 0$, so its only real root is $x \approx 2.09$.

A rough check: $f(2) = 8 - 4 - 5 = -1 < 0$ and $f(3) = 27 - 6 - 5 = 16 > 0$, so $f$ changes sign between $2$ and $3$, as expected.

(a) Using the minimum and maximum functions:

• local minimum points $(-1.47, -5.44)$ and $(1.35, -2.62)$

• local maximum point $(0.126, 0.0627)$

(b) Both ends of the graph rise without bound, so the lowest point overall is the lower of the two local minima. The global minimum value is $-5.44$ (3 s.f.), when $x = -1.47$.

(c) $f$ takes every value from its global minimum upwards, so the range is $f(x) \ge -5.44$.

(b) $x = 0.5$ is $1.5$ units to the left of $x = 2$, so its mirror image is $x = 2 + 1.5 = 3.5$. Hence $3.5$ is also a zero.

(c) The vertex of a quadratic lies on its axis of symmetry, so its $x$-coordinate is $2$.

The values approach $2$ in both directions, so the horizontal asymptote is $y = 2$.

(c) $f(0) = \frac{1}{-1} = -1$, so the $y$-intercept is $(0, -1)$. Solving $f(x) = 0$ means the numerator is zero: $2x + 1 = 0$, so the $x$-intercept is $(-0.5, 0)$.

• y = x²
• y = 2ˣ

(c) For $k > 2$ or $k < -2$ the line meets only one branch: exactly one solution.

(c) Compare the turning values with the endpoints: $f(0) = 1$ and $f(4.5) = 91.125 - 121.5 + 40.5 + 1 = 11.125$.

• Global maximum value: $11.125$, at the endpoint $x = 4.5$ (larger than the local maximum value $5$).

• Global minimum value: $1$, at $x = 0$ and at $x = 3$.

(d) The range is $1 \le f(x) \le 11.125$.

(e) The graph rises from $(0, 1)$ to $(1, 5)$, falls to $(3, 1)$, then rises to $(4.5, 11.125)$. A horizontal line $y = k$ meets all three sections when $1 < k < 5$.

At $k = 1$ there are only two solutions ($x = 0$ and $x = 3$), and at $k = 5$ only two ($x = 1$, where the line touches the maximum, and $x = 4$, since $f(4) = 64 - 96 + 36 + 1 = 5$). So exactly three solutions for $1 < k < 5$.

$5.58$ hours after 6 a.m. is about 11:35 a.m. ($0.5826 \times 60 \approx 35$ minutes), and $t = 10$ is 4 p.m.

(c) Solve $T(t) = 18$ with the GDC: $t = 2.5151\ldots$ and $t = 11.5503\ldots$. The temperature is at least $18\,°\text{C}$ between these times, so the fan runs for

$(q - \cos\theta)^2 = 4 - \sin^2\theta$ A1

$q - \cos\theta > 0$, so the positive square root is taken R1

$q = \cos\theta + \sqrt{4 - \sin^2\theta}$ AG

$= \dfrac{\sqrt{15} - \sqrt{3}}{2}$ $\;(1.07046\ldots)$ A1

OR

$4 - \sin^2\theta = (2.5 - \cos\theta)^2 \Rightarrow 5\cos\theta = 3.25$ (M1)

THEN

$\theta = 0.863$ $\;(0.863211\ldots)$ A1

$\theta = 5.42$ $\;(5.41997\ldots)$ A1

[3]3 marks

$\theta = \dfrac{\pi}{2}$ A1