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Class 9 Mathematics Notes – Chapter 10: Graphs of Functions

Class: 9th
Subject: Mathematics
Board: Punjab Board
Chapter: 10 – Graphs of Functions
Website: QuranEducationPoint.com

Exercise 10.1

Q1. Sketch the graph of the following linear functions.

(i) \(y=3x-5\)

Step 1: Find the intercepts

– \(y\)-intercept: Put \(x=0\).

\[ y=3(0)-5=-5 \]

Therefore, the \(y\)-intercept is \((0,-5)\).

– \(x\)-intercept: Put \(y=0\).

\[ 0=3x-5 \]

\[ 3x=5 \]

\[ x=\frac{5}{3}\approx1.67 \]

Therefore, the \(x\)-intercept is approximately \((1.67,0)\).

Step 2: Table of values

\(x\)	-1	0	2
\(y\)	-8	-5	1

Step 3: Graph characteristics

Plot the points \((0,-5)\), \((1.67,0)\), and \((2,1)\). Connect them with a straight line rising to the right with slope \(m=3\).

(ii) \(y=-2x+8\)

Step 1: Find the intercepts

– \(y\)-intercept: Put \(x=0\).

\[ y=-2(0)+8=8 \]

Therefore, the \(y\)-intercept is \((0,8)\).

– \(x\)-intercept: Put \(y=0\).

\[ 0=-2x+8 \]

\[ 2x=8 \]

\[ x=4 \]

Therefore, the \(x\)-intercept is \((4,0)\).

Step 2: Table of values

\(x\)	0	2	4
\(y\)	8	4	0

Step 3: Graph characteristics

Plot the points \((0,8)\), \((2,4)\), and \((4,0)\). Connect them with a straight line falling to the right with slope \(m=-2\).

(iii) \(y=0.5x-1\)

Step 1: Find the intercepts

– \(y\)-intercept: Put \(x=0\).

\[ y=0.5(0)-1=-1 \]

Therefore, the \(y\)-intercept is \((0,-1)\).

– \(x\)-intercept: Put \(y=0\).

\[ 0=0.5x-1 \]

\[ 0.5x=1 \]

\[ x=2 \]

Therefore, the \(x\)-intercept is \((2,0)\).

Step 2: Table of values

\(x\)	-2	0	4
\(y\)	-2	-1	1

Step 3: Graph characteristics

Plot the points \((-2,-2)\), \((0,-1)\), and \((2,0)\). Draw a straight line through them.

Q2. Plot the graphs of the following quadratic and cubic functions.

(i) \(y=x^3+2x^2-5x-6\), for \(-3.5\leq x\leq2.5\)

Step 1: Calculate the values

– When \(x=-3.5\):

\[ \begin{aligned} y&=(-3.5)^3+2(-3.5)^2-5(-3.5)-6\\ &=-42.875+24.5+17.5-6\\ &=-6.875 \end{aligned} \]

– When \(x=-3\):

\[ y=-27+18+15-6=0 \]

– When \(x=-2\):

\[ y=-8+8+10-6=4 \]

– When \(x=-1\):

\[ y=-1+2+5-6=0 \]

– When \(x=0\):

\[ y=-6 \]

– When \(x=1\):

\[ y=1+2-5-6=-8 \]

– When \(x=2\):

\[ y=8+8-10-6=0 \]

– When \(x=2.5\):

\[ \begin{aligned} y&=(2.5)^3+2(2.5)^2-5(2.5)-6\\ &=15.625+12.5-12.5-6\\ &=9.625 \end{aligned} \]

Step 2: Summary table

\(x\)	-3.5	-3	-2	-1	0	1	2	2.5
\(y\)	-6.88	0	4	0	-6	-8	0	9.63

Step 3: Plotting

Plot the coordinates and draw a smooth S-shaped cubic curve passing through the \(x\)-intercepts at \(x=-3,-1,2\).

(ii) \(y=x^2+x-2\)

Step 1: Table of values

\(x\)	-3	-2	-0.5	0	1	2
\(y\)	4	0	-2.25	-2	0	4

Step 2: Key features

– Vertex:

\[ x=-\frac{b}{2a}=-\frac{1}{2}=-0.5 \]

Substituting \(x=-0.5\):

\[ y=(-0.5)^2-0.5-2=-2.25 \]

Therefore, the vertex is \((-0.5,-2.25)\).

\(x\)-intercepts: \(x=-2\) and \(x=1\).
\(y\)-intercept: \((0,-2)\).

Step 3: Plotting

Draw a smooth, upward-opening parabolic curve through the plotted points.

(iii) \(y=x^3+3x^2+2x\), for \(-2.5\leq x\leq0.5\)

Step 1: Table of values

– When \(x=-2.5\):

\[ y=(-2.5)^3+3(-2.5)^2+2(-2.5) \]

\[ y=-15.625+18.75-5=-1.875 \]

– When \(x=-2\):

\[ y=-8+12-4=0 \]

– When \(x=-1.5\):

\[ y=-3.375+6.75-3=0.375 \]

– When \(x=-1\):

\[ y=-1+3-2=0 \]

– When \(x=-0.5\):

\[ y=-0.125+0.75-1=-0.375 \]

– When \(x=0\):

\[ y=0 \]

– When \(x=0.5\):

\[ y=0.125+0.75+1=1.875 \]

\(x\)	-2.5	-2	-1.5	-1	-0.5	0	0.5
\(y\)	-1.875	0	0.375	0	-0.375	0	1.875

Step 2: Plotting

Plot the points and draw a smooth cubic curve. The curve crosses the \(x\)-axis at \(x=-2,-1,0\).

(iv) \(y=5x^2-2x-3\)

Step 1: Table of values

\(x\)	-1	0	0.2	1	2
\(y\)	4	-3	-3.2	0	13

Step 2: Plotting

Draw a U-shaped parabola opening upwards with \(x\)-intercepts at \(x=1\) and \(x=-0.6\).

Q3. Plot the graph of the following functions.

(i) \(y=4^x\)

Type: Exponential growth function.

Table of values

\(x\)	-2	-1	0	1	2
\(y\)	0.0625	0.25	1	4	16

Plotting: The curve passes through \((0,1)\), increases rapidly for \(x>0\), and approaches \(y=0\) as \(x\) becomes negative.

(ii) \(y=5^{-x}\)

Type: Exponential decay function.

Table of values

\(x\)	-2	-1	0	1	2
\(y\)	25	5	1	0.2	0.04

Plotting: The curve falls from left to right and passes through \((0,1)\).

(iii) \(y=\frac{1}{x-3}\), \(x\neq3\)

Type: Reciprocal function with asymptotes.

Vertical asymptote: \(x=3\)
Horizontal asymptote: \(y=0\)

Table of values

\(x\)	1	2	2.5	3.5	4	5
\(y\)	-0.5	-1	-2	2	1	0.5

Plotting: Draw two branches of a rectangular hyperbola, one on each side of the vertical asymptote \(x=3\).

(iv) \(y=\frac{2}{x}+3\), \(x\neq0\)

Type: Reciprocal function.

Vertical asymptote: \(x=0\), the \(y\)-axis.
Horizontal asymptote: \(y=3\).

Table of values

\(x\)	-2	-1	-0.5	0.5	1	2
\(y\)	2	1	-1	7	5	4

(v) \(y=x^{1/2}=\sqrt{x}\), \(x\geq0\)

Type: Radical or power function.

Table of values

\(x\)	0	1	4	9	16
\(y\)	0	1	2	3	4

Plotting: The curve starts at the origin \((0,0)\) and curves smoothly into the first quadrant.

(vi) \(y=3x^{1/3}=3\sqrt[3]{x}\)

Type: Cube-root power function.

Table of values

\(x\)	-8	-1	0	1	8
\(y\)	-6	-3	0	3	6

Plotting: Draw a smooth S-shaped curve passing through the origin \((0,0)\).

(vii) \(y=2x^{-2}=\frac{2}{x^2}\), \(x\neq0\)

Type: Reciprocal power function.

Table of values

\(x\)	-2	-1	-0.5	0.5	1	2
\(y\)	0.5	2	8	8	2	0.5

Plotting: Draw a curve symmetric about the \(y\)-axis, with both branches above the \(x\)-axis.

Exercise 10.2

Q1. Plot \(y=2x^2-4x+3\) from \(x=-1\) to \(x=3\). Draw a tangent at \((2,3)\) and find its gradient.

Step 1: Calculate the table of values

\(x\)	-1	0	1	2	3
\(y\)	9	3	1	3	9

Step 2: Draw the curve

Plot the points \((-1,9)\), \((0,3)\), \((1,1)\), \((2,3)\), and \((3,9)\), and join them with a smooth parabola.

Step 3: Find the gradient at \((2,3)\)

The derivative is:

\[ \frac{dy}{dx}=4x-4 \]

At \(x=2\):

\[ \text{Gradient}=4(2)-4=4 \]

Answer: Gradient = 4.

Q2. Plot \(y=3x^2+x+1\) and draw a tangent at \((1,5)\). Find its gradient.

Step 1: Table of values

\(x\)	-2	-1	0	1	2
\(y\)	11	3	1	5	15

Step 2: Find the gradient at \((1,5)\)

The derivative is:

\[ \frac{dy}{dx}=6x+1 \]

At \(x=1\):

\[ \text{Gradient}=6(1)+1=7 \]

Answer: Gradient = 7.

Q3. Student strength equation: \(S=1000e^{-t}\)

(a) Graph the equation for \(t=0\) (2016) to \(t=9\) (2025).

\(t\)	0 (2016)	3 (2019)	7 (2023)	9 (2025)
\(S=1000e^{-t}\)	1000	Approximately 50	Approximately 1	Approximately 0

(b) Estimate the student strength in 2019 and 2023.

In 2019, \(t=3\):

\[ S=1000e^{-3}=1000(0.049787)\approx50 \]

Answer: Approximately 50 students.

In 2023, \(t=7\):

\[ S=1000e^{-7}=1000(0.000912)\approx1 \]

Answer: Approximately 1 student.

Q4. Demand and supply functions

Given:

\[ P_d=400-5Q \]

\[ P_s=3Q+24 \]

For \(Q=0\) to \(300\):

Step 1: Table of values

\(Q\)	0	47	100	300
\(P_d=400-5Q\)	400	165	-100	-1100
\(P_s=3Q+24\)	24	165	324	924

Step 2: Find the market equilibrium point

At equilibrium:

\[ P_d=P_s \]

\[ 400-5Q=3Q+24 \]

\[ 8Q=376 \]

\[ Q=47 \]

Substitute \(Q=47\):

\[ P=400-5(47)=165 \]

Answer: Equilibrium point is \((Q,P)=(47,165)\).

Q5. Salary function: \(S(x)=45000+4500x\), for \(0\leq x\leq5\)

Step 1: Table of values

\(x\) (years)	0	1	2	3	4	5
\(S(x)\) (Rs.)	45,000	49,500	54,000	58,500	63,000	67,500

Step 2: Interpretation

The graph is a straight line with an initial salary of Rs. 45,000 at \(x=0\) and a yearly increase of Rs. 4,500.

Q6. School bags production

Given:

\[ C(x)=1200+800x \]

\[ R(x)=2500x \]

(a) Find the break-even point.

At break-even:

\[ R(x)=C(x) \]

\[ 2500x=1200+800x \]

\[ 1700x=1200 \]

\[ x=\frac{1200}{1700}\approx0.7059 \]

Answer: Break-even occurs at approximately \(0.706\) bags according to the given equations.

(b) Find the profit or loss when 250 bags are sold.

Revenue:

\[ R(250)=2500(250)=625000 \]

Cost:

\[ C(250)=1200+800(250) \]

\[ C(250)=1200+200000=201200 \]

Profit:

\[ \begin{aligned} \text{Profit}&=R(250)-C(250)\\ &=625000-201200\\ &=423800 \end{aligned} \]

Answer: Profit = Rs. 423,800.

(c) Plot the graphs.

Plot both straight lines \(C(x)\) and \(R(x)\) on the same coordinate plane to show their intersection.

Q7. Newspaper agency profit

Given:

\[ p(x)=0.10x+70 \]

Step 1: Table of values

\(x\) (newspapers)	0	250	500
\(p(x)\) (Rs.)	70	95	120

Step 2: Find the profit for 500 newspapers.

\[ p(500)=0.10(500)+70 \]

\[ =50+70=120 \]

Answer: Rs. 120.

Q8. Shirt cost function

Given:

\[ C(x)=1500+10x+0.2x^2 \]

for \(0\leq x\leq150\).

Find the cost of 200 shirts by evaluating the formula at \(x=200\):

\[ C(200)=1500+10(200)+0.2(200)^2 \]

\[ =1500+2000+0.2(40000) \]

\[ =1500+2000+8000 \]

Answer: \(C(200)=\text{Rs. }11,500\).

Review Exercise 10

Q1. Multiple-choice questions

(i) \(x=5\) represents:

Correct Answer: (d) Line parallel to the \(y\)-axis.

(ii) The slope of the line \(y=5x+3\) is:

Correct Answer: (c) 5.

(iii) The \(y\)-intercept of \(y=-2x-1\) is:

Correct Answer: (c) -1.

(iv) The graph of \(y=x^3\) cuts the \(x\)-axis at:

Correct Answer: (a) \(x=0\).

(v) The graph of \(3^x\) represents:

Correct Answer: (a) Growth.

(vi) The graph of \(y=-x^2+5\) opens:

Correct Answer: (b) Downward.

(vii) The graph of \(y=x^2-9\) opens:

Correct Answer: (a) Upward.

(viii) \(y=5^x\) is a ______ function.

Correct Answer: (d) Exponential.

(ix) A reciprocal function is:

Correct Answer: (b) \(y=\frac{2}{x}\).

(x) \(y=-3x^3+7\) is a ______ function.

Correct Answer: (b) Cubic.

Q2. Plot the graphs of the following functions.

(i) \(y=3^{-x}\), for \(x\) from \(-2\) to \(4\)

Table of values

\(x\)	-2	-1	0	1	2	3	4
\(y\)	9	3	1	0.33	0.11	0.037	0.012

Plotting: Draw an exponential decay curve passing through \((0,1)\).

(ii) \(y=\frac{2}{x+7}\), \(x\neq-7\)

Type: Reciprocal function.

Vertical asymptote: \(x=-7\)
Horizontal asymptote: \(y=0\)

Table of values

\(x\)	-9	-8	-7.5	-6.5	-6	-5
\(y\)	-1	-2	-4	4	2	1

Q3. Magazine sales equation

Given:

\[ S=200000(1-e^{-0.05t}) \]

(a) Plot the graph for the first 50 weeks.

\(t\) (weeks)	0	5	20	35	50
\(S\)	0	44,240	126,424	165,244	183,583

(b) Calculate the number of magazines sold when \(t=5\) and \(t=35\).

For \(t=5\):

\[ S=200000(1-e^{-0.25}) \]

\[ =200000(1-0.7788)=44240 \]

Answer: 44,240 magazines.

For \(t=35\):

\[ S=200000(1-e^{-1.75}) \]

\[ =200000(1-0.17377)\approx165246 \]

Answer: Approximately 165,246 magazines.

Q4. Plot the graphs for \(x\) from \(-5\) to \(5\).

(i) \(y=x^2-3\)

Table of values

\(x\)	-5	-3	0	3	5
\(y\)	22	6	-3	6	22

Graph characteristics: An upward-opening parabola with vertex at \((0,-3)\).

(ii) \(y=15-x^2\)

Table of values

\(x\)	-5	-3	0	3	5
\(y\)	-10	6	15	6	-10

Graph characteristics: A downward-opening parabola with vertex at \((0,15)\).

Q5. Plot the graph of

\[ y=\frac12(x+4)(x-1)(x-3) \]

from \(x=-5\) to \(x=4\).

Step 1: Calculate the values

– When \(x=-4\):

\[ y=0 \]

– When \(x=-2\):

\[ y=\frac12(2)(-3)(-5)=15 \]

– When \(x=-1\):

\[ y=\frac12(3)(-2)(-4)=12 \]

– When \(x=0\):

\[ y=\frac12(4)(-1)(-3)=6 \]

– When \(x=1\):

\[ y=0 \]

– When \(x=2\):

\[ y=\frac12(6)(1)(-1)=-3 \]

– When \(x=3\):

\[ y=0 \]

– When \(x=4\):

\[ y=\frac12(8)(3)(1)=12 \]

Step 2: Table of values

\(x\)	-4	-2	-1	0	1	2	3	4
\(y\)	0	15	12	6	0	-3	0	12

Step 3: Plotting

Plot the points and draw a smooth cubic curve crossing the \(x\)-axis at \(x=-4,1,3\).

Q6. Market functions

Given:

\[ P_s=Q^2+5 \]

\[ P_d=Q^2-10Q \]

for \(Q=-20\) to \(20\).

Table of values

\(Q\)	-20	-10	0	10	20
\(P_s=Q^2+5\)	405	105	5	105	405
\(P_d=Q^2-10Q\)	600	200	0	0	200

Plotting: Plot both quadratic curves on the same graph.

Q7. LED manufacturer

Given:

\[ C(x)=60000+250x \]

\[ R(x)=1200x \]

(a) Find the break-even point.

At break-even:

\[ R(x)=C(x) \]

\[ 1200x=60000+250x \]

\[ 950x=60000 \]

\[ x=\frac{60000}{950}\approx63.16 \]

Answer: The break-even point is approximately 63.16 LEDs, or about 63–64 LEDs.

(b) Find the profit or loss when 100 LEDs are sold.

Revenue:

\[ R(100)=1200(100)=120000 \]

Cost:

\[ C(100)=60000+250(100) \]

\[ C(100)=60000+25000=85000 \]

Profit:

\[ \begin{aligned} \text{Profit}&=R(100)-C(100)\\ &=120000-85000\\ &=35000 \end{aligned} \]

Answer: Profit = Rs. 35,000.

Notes by Sir Dilawar

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