Quadratic Graphs (Cambridge International IGCSE Mathematics)

Quadratic graphs

What is a quadratic graph?

• A quadratic graph has the form $y=ax^{2}+bx+c$

  • 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 $x^{2}$ gives a u-shaped curve
  • A negative number in front of $x^{2}$ 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 intersects the$x$-axis twice, once, or not at all

  • The x values where the graph intersects 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 turning points

  • A turning point can also be called a vertex

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 $y=ax^{2}+bx+c$ 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 $y=0$; $ax^{2}+bx+c=0$
    • You can find the solutions by factorising, completing the square or using the quadratic formula

  • Identify if the number $a$ in $ax^{2}+bx+c$ 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

How do I find the coordinates of the turning point by completing the square?

• The coordinates of the turning point(vertex) of a quadratic graph can be found by completing the square

• For a quadratic graph written in the form $y = a ( x - p )^{2} + q$

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

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

  • A curve with equation $y = ( x - 3 )^{2} + 2$, 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 turning point but it will change the shape of the graph

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

    • The curve $y = 5 ( x - 3 )^{2} + 2$ has a minimum point at $( 3 , 2 )$

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

    • The curve $y = - 8 ( x - 3 )^{2} + 2$ has a maximum point at $( 3 , 2 )$

How do I find the coordinates of the turning point using differentiation?

• The coordinates of the turning point(maximum/minimum) of a quadratic can be found through differentiation

• To find the coordinates of the turning point

  • Differentiate the quadratic equation $y=ax^{2}+bx+c$

    • This will give you $\frac{dy}{dx}$

  • Set $\frac{dy}{dx}=0$ and solve for $x$

    • The solution will be the $x$-coordinate of the turning point

  • Substitute the $x$ value into $y=ax^{2}+bx+c$

    • This will give you the $y$-coordinate of the turning point

Worked Example (a) Sketch the graph of $y=x^{2}-5x+6$ showing the $x$ and $y$ intercepts clearly.

Answer:

The $+c$ at the end is the $y$-intercept

$y$-intercept: (0, 6)

Factorise the quadratic expression

$y = ( x - 2 ) ( x - 3 )$

Solve $y=0$

$( x - 2 ) ( x - 3 ) = 0 , \text{so} x = 2 \text{or} x = 3$

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

(b) Sketch the graph of $y=x^{2}-6x+13$ showing the $y$-intercept and the coordinates of the turning point.

Answer:

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

The $+c$ at the end is the $y$-intercept

$y$-intercept: (0, 13)

Find the minimum point by completing the square (or through differentiation)

For example, complete the square by writing the equation in the form $a ( x - p )^{2} + q$ (you may need to look this method up)

$\begin{gathered} x^{2} - 6 x + 13 & = ( x - 3 )^{2} - 9 + 13 \\ = ( x - 3 )^{2} + 4 \end{gathered}$

The turning point of $y = a ( x - p )^{2} + q$ has coordinates $( p , q )$ The minimum point is therefore

(3, 4)

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)

(c) Sketch the graph of $y=-x^{2}-4x-4$ showing the root(s), $y$-intercept, and the coordinates of the turning point.

Answer:

It is a negative quadratic, so will be an n-shape The turning point will therefore be a maximum

The $+c$ at the end is the $y$-intercept

• $y$-intercept: (0, -4)*

Find the minimum point by completing the square or through differentiation

For example, differentiate the equation

$\frac{dy}{dx}=-2x-4$

Set the derivative equal to zero and solve for $x$

$\begin{gathered} 0 = - 2 x - 4 \\ 2 x & = - 4 \\ x = - 2 \end{gathered}$

Substitute this value of $x$ back into the original equation for $y$

$\begin{gathered} y = - ( - 2 )^{2} - 4 ( - 2 ) - 4 \\ y = - 4 + 8 - 4 \\ y & = 0 \end{gathered}$

This shows that the maximum point has coordinates

(-2, 0)

As the maximum is on the $x$*-axis, there is *only one root

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

• If the vertex and one other point are known

  • Use the form $y = a ( x - p )^{2} + q$ 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 $y = a ( x - x_{1} ) ( x - x_{2} )$ to fill in $x_{1}$ and $x_{2}$

    • The roots are at $( x_{1} , 0 )$ and $( x_{2} , 0 )$

  • 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.

Answer:

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

Use the form $y = a ( x - x_{1} ) ( x - x_{2} )$ to fill in $x_{1}$ and $x_{2}$ by inspection

The roots are at $( x_{1} , 0 )$ and $( x_{2} , 0 )$

$y = a ( x - 2 ) ( x - 3 )$

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

$\begin{gathered} 24 = a ( 0 - 2 ) ( 0 - 3 ) \\ 24 = a ( - 2 ) ( - 3 ) \\ 24 = 6 a \\ 4 = a \end{gathered}$

Write the full equation

$y=4(x-2)(x-3)$

You could also write this in expanded form: $y=4x^{2}-20x+24$

(b) Find the equation of the graph below.

Answer:

The graph shows the vertex and a point on the curve

Use the form $y = a ( x - p )^{2} + q$ to fill in $p$ and $q$ by inspection

The vertex is at $( p , q )$

$y = a ( x - 9 )^{2} - 16$

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

$\begin{gathered} 82 = a ( 2 - 9 )^{2} - 16 \\ 82 = a ( - 7 )^{2} - 16 \\ 82 = 49 a - 16 \\ 98 = 49 a \\ 2 = a \end{gathered}$

Write the full equation

$y=2(x-9)^{2}-16$or *$y=2x^{2}-36x+146$