Studying and graphing a function

Lan wants to draw the graph of $y=x^3-3x+1$ by hand. She plots five points:

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$-1$	$3$	$1$	$-1$	$3$

and joins them up. But what if, between two plotted points, the graph has a "bump" that Lan cannot see? And how many solutions does the equation $x^3-3x+1=0$ have?

The curve-sketching procedure

Explore: the derivative tells the shape

The derivative $y'=3x^2-3=3(x-1)(x+1)$ is zero only at $x=-1$ and $x=1$ . So the graph can only "turn round" at those two points: it goes up on $(-\infty, -1)$ , down on $(-1, 1)$ and up on $(1, +\infty)$ . There is no other bump between Lan's points.

So a few points and the table of variations are enough to draw the graph with the right shape. That is the idea of the procedure.

Cubic functions

Example Analysing

y=x^3-3x+1

Analyse $y=x^3-3x+1$ and sketch its graph. How many solutions does $x^3-3x+1=0$ have?

Solution. The domain is $\mathbb{R}$ . We compute the derivative and the limits at infinity:

$y'=3(x-1)(x+1)=0\iff x=\pm1;\qquad \lim_{x\to+\infty}y=+\infty,\quad \lim_{x\to-\infty}y=-\infty.$

Substituting $x=\pm1$ : $y(-1)=3$ , $y(1)=-1$ . The table of variations:

$x$	$-\infty$		$-1$		$1$		$+\infty$
$y'$		$+$	$0$	$-$	$0$	$+$
$y$	$-\infty$	$\nearrow$	$3$	$\searrow$	$-1$	$\nearrow$	$+\infty$

The function is increasing on $(-\infty, -1)$ and $(1, +\infty)$ and decreasing on $(-1, 1)$ ; local maximum $(-1, 3)$ , local minimum $(1, -1)$ . The graph meets the y-axis at $(0, 1)$ .

To count the solutions, we count how many times the graph crosses the x-axis. On each of $(-\infty, -1)$ , $(-1, 1)$ , $(1, +\infty)$ the function is monotonic and $y$ changes sign exactly once (from $-\infty$ up to $3$ , from $3$ down to $-1$ , from $-1$ up to $+\infty$ ).

So the equation $x^3-3x+1=0$ has exactly $3$ solutions (approximately $-1.88$ , $0.35$ , $1.53$ ).

The graph of

y=x^3-3x+1

The point $(0, 1)$ is the midpoint of the two turning points, and the graph is symmetric about it. Every cubic graph has a centre of symmetry: the point whose x-coordinate is the root of $y''=0$ (here $y''=6x=0\iff x=0$ ).

Not every cubic has extrema. When $y'$ does not change sign, the graph goes one way only.

Example A cubic with no extrema

Analyse $y=-x^3+3x^2-3x+2$ and sketch its graph.

Solution. The domain is $\mathbb{R}$ . The derivative is a negative number times a square:

$y'=-3x^2+6x-3=-3(x-1)^2\le0,\qquad y'=0\iff x=1.$

Limits: $\lim\limits_{x\to+\infty}y=-\infty$ , $\lim\limits_{x\to-\infty}y=+\infty$ . The derivative is zero at only one point, so the function is decreasing on $\mathbb{R}$ and has no extremum. We have $y''=-6x+6=0\iff x=1$ , $y(1)=1$ : the centre of symmetry is $I(1, 1)$ , where the tangent is horizontal.

Intercepts: $y(0)=2$ ; and $-x^3+3x^2-3x+2=-(x-2)(x^2-x+1)=0\iff x=2$ (since $x^2-x+1$ has $\Delta=-3<0$ ).

So the function is decreasing on $\mathbb{R}$ ; the graph passes through $(0, 2)$ , $I(1, 1)$ and $(2, 0)$ .

The graph of

y=-x^3+3x^2-3x+2

Rational functions

For a rational function, the step "limits and asymptotes" matters more: we draw the asymptotes first as dashed lines, then draw the branches close to them.

Example Analysing

y=\dfrac{x+1}{x-1}

Analyse $y=\dfrac{x+1}{x-1}$ and sketch its graph.

Solution. The domain is $D=\mathbb{R}\setminus\{1\}$ . We use the formula $\left(\dfrac{ax+b}{cx+d}\right)'=\dfrac{ad-bc}{(cx+d)^2}$ :

$y'=\frac{1\cdot(-1)-1\cdot1}{(x-1)^2}=\frac{-2}{(x-1)^2}<0\quad\text{for every } x\in D.$

Asymptotes: $\lim\limits_{x\to\pm\infty}y=1$ , so $y=1$ is a horizontal asymptote; $\lim\limits_{x\to1^+}y=+\infty$ , $\lim\limits_{x\to1^-}y=-\infty$ , so $x=1$ is a vertical asymptote.

$x$	$-\infty$		$1$		$+\infty$
$y'$		$-$	$\Vert$	$-$
$y$	$1$	$\searrow$	$-\infty\ \Vert\ +\infty$	$\searrow$	$1$

Intercepts: $(-1, 0)$ and $(0, -1)$ .

So the function is decreasing on each interval $(-\infty, 1)$ and $(1, +\infty)$ and has no extremum; the graph is symmetric about the intersection $I(1, 1)$ of the two asymptotes.

The graph of

y=\dfrac{x+1}{x-1}

For $y=\dfrac{ax^2+bx+c}{mx+n}$ , we divide first: we see the oblique asymptote, and the derivative becomes easier.

Example Analysing

y=\dfrac{x^2-2x+2}{x-1}

Analyse $y=\dfrac{x^2-2x+2}{x-1}$ and sketch its graph.

Solution. The domain is $D=\mathbb{R}\setminus\{1\}$ . Dividing: $x^2-2x+2=(x-1)(x-1)+1$ , so

$y=x-1+\frac{1}{x-1},\qquad y'=1-\frac{1}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}.$

$y'=0\iff x=0$ or $x=2$ ; $y(0)=-2$ , $y(2)=2$ . Asymptotes: vertical $x=1$ (since $\lim\limits_{x\to1^+}y=+\infty$ ), oblique $y=x-1$ (since $\dfrac{1}{x-1}\to0$ ).

$x$	$-\infty$		$0$		$1$		$2$		$+\infty$
$y'$		$+$	$0$	$-$	$\Vert$	$-$	$0$	$+$
$y$	$-\infty$	$\nearrow$	$-2$	$\searrow$	$-\infty\ \Vert\ +\infty$	$\searrow$	$2$	$\nearrow$	$+\infty$

The graph meets the y-axis at $(0, -2)$ and does not meet the x-axis, because $x^2-2x+2$ has $\Delta'=1-2<0$ .

So the function has the local maximum $(0, -2)$ , the local minimum $(2, 2)$ , the vertical asymptote $x=1$ and the oblique asymptote $y=x-1$ .

The graph of

y=\dfrac{x^2-2x+2}{x-1}

Reading graphs and applications

Once we have the graph, we can read a lot without computing again, for example the number of solutions of $f(x)=k$ (for a given number $k$ ): it is the number of intersection points of the graph with the horizontal line $y=k$ .

The graph of a cubic

y=f(x)

and the line

y=1

Example Reading a graph

The figure above is the graph of $f(x)=ax^3+bx^2+cx+d$ , with local maximum point $(0, 4)$ and local minimum point $(2, 0)$ . a) Find $a$ , $b$ , $c$ , $d$ . b) How many solutions does $f(x)=1$ have? What about $f(x)=4$ ?

Solution. a) We use four facts: $f(0)=4$ , $f'(0)=0$ , $f'(2)=0$ , $f(2)=0$ , with $f'(x)=3ax^2+2bx+c$ :

$\begin{aligned} f(0)=4 &\Rightarrow d=4; & f'(0)=0 &\Rightarrow c=0;\\ f'(2)=0 &\Rightarrow 12a+4b=0\Rightarrow b=-3a; & f(2)=0 &\Rightarrow 8a-12a+4=0\Rightarrow a=1. \end{aligned}$

So $f(x)=x^3-3x^2+4$ (then $b=-3$ ).

b) The line $y=1$ lies between the local minimum value $0$ and the local maximum value $4$ , so it cuts the graph at $3$ points. The line $y=4$ passes through the local maximum point (touching the graph there) and cuts the right-hand part at one more point: indeed $x^3-3x^2+4=4\iff x^2(x-3)=0$ .

So $f(x)=x^3-3x^2+4$ ; the equation $f(x)=1$ has $3$ solutions, and $f(x)=4$ has $2$ solutions ( $x=0$ and $x=3$ ).

Finally, we recall how to write a tangent line from grade 11: the tangent at $M(x_0, y_0)$ is $y=f'(x_0)(x-x_0)+y_0$ .

Watch out Common mistakes

Drawing a rational graph in one stroke. Wrong: joining the two branches of $y=\dfrac{x+1}{x-1}$ across $x=1$ . Why it is wrong: the function is undefined at $x=1$ , and the vertical asymptote splits the graph into two branches. Right: draw the dashed asymptotes first, with one branch on each side.
Miscounting when the line passes through a turning point. Wrong: " $f(x)=4$ has $3$ solutions because $4$ is within the range of the function". Why it is wrong: $y=4$ only touches the graph at the local maximum and does not cross it twice there. Right: $2$ solutions.
Not dividing, so missing the oblique asymptote. Wrong: " $y=\dfrac{x^2-2x+2}{x-1}$ only has a vertical asymptote". Why it is wrong: $y=x-1+\dfrac{1}{x-1}$ . Right: there is also the oblique asymptote $y=x-1$ .
Using $y'(x)$ instead of $y'(x_0)$ in a tangent. Wrong: $y=\dfrac{-2}{(x-1)^2}(x-2)+3$ . Why it is wrong: the gradient is a number. Right: $y=-2(x-2)+3$ .

Key terms

English	Tiếng Việt
curve sketching (analysing a function)	khảo sát hàm số
curve-sketching procedure	sơ đồ khảo sát
table of variations	bảng biến thiên
centre of symmetry	tâm đối xứng
intersection point	giao điểm
tangent line	tiếp tuyến
gradient	hệ số góc
number of solutions of an equation	số nghiệm của phương trình

Summary

Procedure: domain → $y'$ , limits, asymptotes, table of variations → intercepts, special points → draw.
A cubic has either two extrema or none; its centre of symmetry is the point where $y''=0$ .
$y=\dfrac{ax+b}{cx+d}$ : always increasing or always decreasing on each interval; its two asymptotes meet at the centre of symmetry.
$y=\dfrac{ax^2+bx+c}{mx+n}$ : divide to get the oblique asymptote and a tidy derivative.
The number of solutions of $f(x)=k$ is the number of intersections of the graph with the line $y=k$ . Tangent at $x_0$ : $y=f'(x_0)(x-x_0)+f(x_0)$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Analyse $y=x^3-3x$ and sketch its graph.

What you need and how to start

What you need: the procedure; an odd function has a graph symmetric about the origin.

Steps:

Compute $y'$ , find the extrema and draw the table of variations.
Find the x-intercepts: $x(x^2-3)=0$ .

Show solution

$y'=3(x-1)(x+1)$ ; increasing on $(-\infty, -1)$ and $(1, +\infty)$ , decreasing on $(-1, 1)$ . Local maximum $(-1, 2)$ , local minimum $(1, -2)$ . The graph meets the x-axis at $x=0$ , $x=\pm\sqrt3$ , and is symmetric about $O$ (also the point where $y''=6x=0$ ).

Answer. Local maximum $(-1, 2)$ , local minimum $(1, -2)$ ; the graph passes through $(-\sqrt3, 0)$ , $O(0, 0)$ , $(\sqrt3, 0)$ , with centre of symmetry $O$ .

Exercise 2

Analyse $y=\dfrac{x-2}{x+1}$ and sketch its graph.

What you need and how to start

What you need: $y=\dfrac{ax+b}{cx+d}$ has $y'=\dfrac{ad-bc}{(cx+d)^2}$ , vertical asymptote $x=-\dfrac dc$ and horizontal asymptote $y=\dfrac ac$ .

Steps:

Compute $y'$ and its sign.
Find the two asymptotes and the intercepts.

Show solution

$D=\mathbb{R}\setminus\{-1\}$ , $y'=\dfrac{1\cdot1-(-2)\cdot1}{(x+1)^2}=\dfrac{3}{(x+1)^2}>0$ : increasing on each interval $(-\infty, -1)$ and $(-1, +\infty)$ . Vertical asymptote $x=-1$ , horizontal asymptote $y=1$ . Intercepts: $(2, 0)$ and $(0, -2)$ .

Answer. Increasing on each interval of the domain; asymptotes $x=-1$ , $y=1$ ; centre of symmetry $I(-1, 1)$ ; the graph passes through $(2, 0)$ and $(0, -2)$ .

Exercise 3

Analyse $y=\dfrac{x^2+x+1}{x+1}$ : find its extrema and asymptotes.

What you need and how to start

What you need: divide to write $y=x+\dfrac{1}{x+1}$ .

Steps:

Compute $y'=1-\dfrac{1}{(x+1)^2}$ and write it as one fraction.
Draw the table of variations with $-2$ , $-1$ , $0$ marked.

Show solution

$y'=\dfrac{x(x+2)}{(x+1)^2}=0\iff x=0$ or $x=-2$ . $y'$ is positive on $(-\infty, -2)$ and $(0, +\infty)$ , negative on $(-2, -1)$ and $(-1, 0)$ . $y(-2)=-3$ , $y(0)=1$ . Vertical asymptote $x=-1$ , oblique asymptote $y=x$ .

Answer. Local maximum $(-2, -3)$ , local minimum $(0, 1)$ ; vertical asymptote $x=-1$ , oblique asymptote $y=x$ .

Exercise 4

Using the graph of $y=x^3-3x$ (exercise 1), find how many solutions each equation has: a) $x^3-3x=1$ ; b) $x^3-3x=-3$ .

What you need and how to start

What you need: the number of solutions of $f(x)=k$ is the number of intersections of the graph with the line $y=k$ .

Steps:

Compare $k$ with the local maximum $2$ and the local minimum $-2$ .
Count the intersections.

Show solution

a) $-2<1<2$ , so the line $y=1$ cuts all three "pieces" of the graph: $3$ solutions. b) $-3<-2$ , so the line $y=-3$ only cuts the left-hand part (where $y$ goes from $-\infty$ up to $2$ ): $1$ solution.

Answer. a) $3$ solutions; b) $1$ solution.

Exercise 5

Consider $y=x^3-6x^2+9x-2$ . Is each claim true or false?

a) The graph meets the y-axis at the point $(0, -2)$ .

b) The local maximum point of the graph is $(1, 2)$ .

c) The equation $x^3-6x^2+9x-2=0$ has exactly two solutions.

d) The centre of symmetry of the graph is the point $(2, 0)$ .

What you need and how to start

What you need: the table of variations of a cubic; the centre of symmetry has the root of $y''=0$ as its x-coordinate.

Steps:

Compute $y'$ and find the extrema.
Compare $0$ with the local maximum and minimum values to count solutions.
Solve $y''=0$ .

Show solution

$y'=3(x-1)(x-3)$ ; $y(1)=2$ (local maximum), $y(3)=-2$ (local minimum).

a) True: $y(0)=-2$ .

b) True: $y'$ changes sign $+\to-$ at $x=1$ and $y(1)=2$ .

c) False: $-2<0<2$ , so the line $y=0$ cuts the graph at $3$ points (the solutions are $2$ and $2\pm\sqrt3$ ).

d) True: $y''=6x-12=0\iff x=2$ , $y(2)=8-24+18-2=0$ .

Answer. a) True; b) True; c) False; d) True.

Exercise 6

The tangent to the graph of $y=x^3-3x^2+4$ at the point with x-coordinate $x_0=3$ meets the y-axis at a point. What is the y-coordinate of that point?

What you need and how to start

What you need: the tangent at $x_0$ is $y=f'(x_0)(x-x_0)+f(x_0)$ (derivatives, grade 11).

Steps:

Compute $f(3)$ and $f'(3)$ .
Write the tangent and put $x=0$ .

Show solution

$f(3)=27-27+4=4$ , and $f'(x)=3x^2-6x$ , so $f'(3)=9$ . Tangent: $y=9(x-3)+4=9x-23$ . For $x=0$ : $y=-23$ .

Answer. $-23$ .

Exercise 7

The number of bacteria in a culture after $t$ hours is $N(t)=\dfrac{500t+100}{t+1}$ (thousand), $t\ge0$ . Analyse the function on $[0, +\infty)$ and say: how many bacteria are there at the start, does the number increase or decrease, and what does it approach in the long run?

What you need and how to start

What you need: the sign of $N'(t)$ tells whether it increases or decreases; the horizontal asymptote gives the long-run value.

Steps:

Compute $N(0)$ and $N'(t)$ .
Compute $\lim\limits_{t\to+\infty}N(t)$ .

Show solution

$N(0)=100$ . $N'(t)=\dfrac{500(t+1)-(500t+100)}{(t+1)^2}=\dfrac{400}{(t+1)^2}>0$ , so $N$ is increasing. $\lim\limits_{t\to+\infty}N(t)=500$ : horizontal asymptote $N=500$ .

Answer. $100$ thousand at the start; the number always increases and approaches $500$ thousand without exceeding it.

Exercise 8

The graph of $y=\dfrac{ax+b}{x+c}$ has the vertical asymptote $x=1$ , the horizontal asymptote $y=2$ , and passes through the point $(0, -1)$ . Find $a$ , $b$ , $c$ .

What you need and how to start

What you need: $y=\dfrac{ax+b}{x+c}$ has the vertical asymptote $x=-c$ and the horizontal asymptote $y=a$ .

Steps:

Deduce $c$ and $a$ from the two asymptotes.
Substitute the point $(0, -1)$ to find $b$ .

Show solution

Vertical asymptote $x=-c=1$ , so $c=-1$ ; horizontal asymptote $y=a=2$ . Substituting $(0, -1)$ : $\dfrac{b}{-1}=-1$ , so $b=1$ . The function is $y=\dfrac{2x+1}{x-1}$ (with $ad-bc=2\cdot(-1)-1\cdot1=-3\neq0$ ).

Answer. $a=2$ , $b=1$ , $c=-1$ .

Exercise 9 Review

Find the coordinates of the intersection points of the graph of $y=\dfrac{x+1}{x-1}$ and the line $y=x-1$ .

What you need and how to start

What you need: the x-coordinates of the intersections solve $\dfrac{x+1}{x-1}=x-1$ with $x\neq1$ ; solve a quadratic equation (grade 10).

Steps:

Multiply both sides by $x-1$ and get a quadratic equation.
Check the condition $x\neq1$ , then compute the y-coordinates.

Show solution

$x+1=(x-1)^2\iff x^2-3x=0\iff x=0$ or $x=3$ (both different from $1$ ). The y-coordinates: $y=0-1=-1$ and $y=3-1=2$ .

Answer. $(0, -1)$ and $(3, 2)$ .

Exercise 10 Challenge

Write the equations of the tangents to the graph of $y=\dfrac{x+1}{x-1}$ that are parallel to the line $y=-2x$ .

What you need and how to start

What you need: parallel lines have the same gradient; the gradient of the tangent at $x_0$ is $y'(x_0)$ .

Steps:

Solve $y'(x_0)=\dfrac{-2}{(x_0-1)^2}=-2$ .
For each root, compute $y_0$ and write the tangent; check that it is not the line $y=-2x$ itself.

Show solution

$(x_0-1)^2=1\iff x_0=0$ or $x_0=2$ . At $x_0=0$ : $y_0=-1$ , tangent $y=-2x-1$ . At $x_0=2$ : $y_0=3$ , tangent $y=-2(x-2)+3=-2x+7$ . Both differ from $y=-2x$ .

Answer. $y=-2x-1$ and $y=-2x+7$ .

The curve-sketching procedure

Explore: the derivative tells the shape

Cubic functions

Rational functions

Reading graphs and applications

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $f$ has a local minimum at $x=1$ .	True	False
b) The centre of symmetry of the graph is $I(2, 1)$ .	True	False
c) The equation $f(x)=3$ has three distinct solutions.	True	False
d) The tangent to the graph at $I(2, 1)$ is $y=3x-5$ .	True	False