Quadratic Graphs (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Quadratic Graphs

What is a quadratic graph?

  • A quadratic graph has the form y=ax2+bx+c 

    • where a is not zero

What does a quadratic graph look like?

  • A quadratic graph is a smooth curve with a vertical line of symmetry

    • A positive number in front of x2 gives a u-shaped curve

    • A negative number in front of x2 gives an n-shaped curve

  • The shape made by a quadratic graph is known as a parabola

  • A quadratic graph will always cross the y-axis

  • A quadratic graph crosses the x-axis twice, once, or not at all

    • The points where the graph crosses the x-axis are called the roots

  • If the graph is a u-shape, it has a minimum point

  • If the graph is an n-shape, it has a maximum point

  • Minimum and maximum points are both examples of a vertex

Diagram showing a positive quadratic curve with a minimum point and a negative quadratic curve with a maximum point.

How do I sketch a quadratic graph?

  • It is important to know how to sketch a quadratic curve

    • A simple drawing showing the key features is often sufficient

    • For a more accurate graph, create a table of values and plot the points

  • To sketch a quadratic graph:

    • First sketch the x and y-axes

    • Identify the y-intercept and mark it on the y-axis

      • The y-intercept of y=ax2+bx+c will be (0, c)

      • It can also be found by substituting in x=0

    • Find all root(s) (0, 1 or 2) of the equation and mark them on the x-axis

      • The roots will be the solutions to y=0ax2+bx+c=0

      • You can find the solutions by factorising or using the quadratic formula

    • Identify whether the number a in ax2+bx+c is positive or negative

      • A positive value will result in a u-shape

      • A negative value will result in an n-shape

    • Sketch a smooth curve through the x and y-intercepts

      • Mark on any axes intercepts

      • Mark on the coordinates of the maximum/minimum point if you know it

Worked Example

Sketch the graph of y=x25x+6 showing the x and y intercepts clearly.

Solution:

The +c at the end is the y-intercept

y-intercept: (0, 6)

Factorise the quadratic expression

y=(x2)(x3)

Solve y=0

(x2)(x3)=0, so x=2 or x=3

So the x-intercepts are given by the coordinates

(2, 0)  and (3, 0)

It is a positive quadratic graph, so will be a u-shape

Graph of y=x²-5x+6 with y-intercept (06) and roots (2, 0), (3, 0) marked on.

How do I find the coordinates of the vertex?

  • For a quadratic graph written in the form y=a(xp)2+q

    • the minimum or maximum point has coordinates (p, q)

  • Beware: there is a sign change for the x-coordinate

    • A curve with equation y=(x3)2+2, has a minimum point at (3, 2)

    • A curve with equation y=(x+3)2+2, has a minimum point at (3, 2)

  • The value of a does not affect the coordinates of the vertex but it will change the shape of the graph

    • If it is positive, a>0, the graph will be a u-shape

      • The curve y=5(x3)2+2 has a minimum point at (3, 2)

    • If it is negative, a<0, the graph will be an n-shape

      • The curve y=8(x3)2+2 has a maximum point at (3, 2)

Worked Example

Sketch the graph of y=(x3)2+4 showing the y-intercept and the coordinates of the vertex.

Answer:

It is a positive quadratic, so will be a u-shape
The vertex will therefore be a minimum

The vertex of y=a(xp)2+q has coordinates (p, q)
The minimum point is therefore

(3, 4)

To find the y-intercept, substitute in x=0

y=(03)2+4y=9+4y=13

y-intercept: (0, 13)

As the minimum point is above the x-axis, and the curve is a u-shape, this means the graph will not cross the x-axis (it has no roots)

Graph of y=x²-6x+13, with y-intercept (0,13) and minimum point (3, 4) marked on.

How do I find the equation of a quadratic from its graph?

  • If the vertex and one other point are known

    • Use the form y=a(xp)2+q to fill in p and q

      • The vertex is at (p, q)

    • Then substitute in the other known point (x, y) to find a

  • If the roots (x-intercepts) and one other point are known

    • Use the form y=a(xx1)(xx2) to fill in x1 and x2

      • The roots are at (x1 , 0) and (x2 , 0)

    • Then substitute in the other known point (x, y) to find a

  • If a=1 then you only need either the vertex or the roots

Worked Example

(a) Find the equation of the graph below.

Positive u-shaped curve with roots at (2,0) and (3,0) and y-intercept of (0,24)

Answer:

The graph shows the roots and a point on the curve (in this case the y-intercept)

Use the form y=a(xx1)(xx2) to fill in x1 and x2 by inspection

The roots are at (x1 , 0) and (x2 , 0)

y=a(x2)(x3)

Substitute in the other known point (0, 24) to find a

24=a(02)(03)24=a(2)(3)24=6a4=a

Write the full equation

y=4(x2)(x3)

You could also write this in expanded form: y=4x220x+24

(b) Find the equation of the graph below.

Positive u-shaped curve with vertex at (9,-16) and a point on the curve at (2,82)

Answer:

The graph shows the vertex and a point on the curve

Use the form y=a(xp)2+q to fill in p and q by inspection

The vertex is at (p, q)

y=a(x9)216

Substitute in the other known point (2, 82) to find a

82=a(29)21682=a(7)21682=49a1698=49a2=a

Write the full equation

y=2(x9)216

You could also write this in expanded form: y=2x236x+146

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