Study S3 Mathematics Quadratic Equations and Functions - Geniebook

Quadratic Equations And Functions

 In this article, we are going to learn about Quadratic Equations and Functions, which cover the following subtopics:

  • Graphs of Quadratic Functions 
    • Determining the shape of a Quadratic Graph 
    • Finding the intercepts of a Quadratic Graph
    • Finding the line of symmetry and the turning point of a Quadratic Graph
  • Sketching Graphs of Quadratic Functions

This article is written to meet the requirements of S3 Mathematics Syllabus in Singapore. 

 

Graphs Of Quadratic Functions

 

Determining The Shape Of A Quadratic Graph

Ideally, the name of the shape of the Quadratic graph is called a “Parabola graph”.

\(y=ax^2+bx+c\)

The given graph above is a minimum point graph.

If the value of \(a>0\), the graph has a minimum point.

\(y=ax^2+bx+c\)

The given graph above is a maximum point graph.

If the value of \(a<0\), the graph has a maximum point.

 

Finding The y-Intercept Of A Quadratic Graph

A quadratic graph will always have only one y-intercept.

 

To find the y-intercept, substitute \(x = 0\) and solve for \(y\).

\(y=ax^2+bx+c\)

Putting in \(x = 0\)

We can find the values of the points where the graph cuts the y-axis.

 

Finding The y-Intercept Of A Quadratic Graph

A quadratic graph may have \(2, 1 \;or\; 0\) x-intercepts.

 

In the first case, as seen, there are two points where the graph cuts off the x-axis.

In the second case, as seen, there is one point where the graph cuts off the x-axis.

In the third case, as seen, there is no point where the graph cuts off the x-axis.

 

To find the x-intercept, substitute \(y = 0\) and solve for \(x\).

\(y=ax^2+bx+c\)

Putting in \(y = 0\),

We can find the values of the points where the graph cuts the x-axis.

 

Finding The Line Of Symmetry Of A Quadratic Graph

The line of symmetry is the line that passes through \(x\) and splits the entire graph into two mirror halves.

Under this, there are 3 different cases:

 

 

Using this simple table below we can find the equation for the line of symmetry

2  x-intercepts

1  x-intercept

No  x-intercepts

\(x=\frac {a+b}{2}\)

\(x=a\)

\(x=a\)

 

Finding The Turning Point Of A Quadratic Graph

A turning point is a point on the graph where it changes its direction. It can either be the turning at the highest point or the lowest point.

For every turning point, the x-coordinate always lies on the line of symmetry. 

 

 

 

Sketching A Quadratic Graph

In the case of a linear equation graph we need to know only the x-intercept and the y-intercept to sketch the graph.

However, in the case of Quadratic graphs, you need a few more parameters to sketch the graph.

For a graph given by the formula \(y=ax^2+bx+c\),

Remember the Acronym S I T.

S

Shape

Coefficient of \(x^2\)

I

Intercepts

The points of \(x, \;y\) coordinates

T

Turning Point

The maximum or minimum turning point

 

 

Graphs Of The Form \(y= (x-h)(x-k)\)

Example: For the graph of \(y= (x-1)(x-3)\) shown above

Shape

Since the coefficient of \(x^2\) is \(1\), hence it is a minimum point graph with an upward parabola

x–intercept

Putting \(y=0\); we get \(x=1\) or \(x=3\)

Line of symmetry

Upon finding the mean of the x-intercepts you get a line of symmetry. \(x =\frac {1+3}{2}=2\)

y–intercept

Putting \(\begin{align*} x=0; \;\;y &= (-1)(-3) =3 \end{align*}\)

 

 

Graphs Of The Form \(y= -(x-h)(x-k)\)

Example: For the graph of \(y= -(x-1)(x-3)\) shown above 

Shape

Since the coefficient of \(x^2\) is \(-1\), hence it is a maximum point graph with downward parabola

x–intercept

Putting \(y=0\); we get \(x=1\) or \(x=3\)

Line of symmetry

Upon finding the mean of the x-intercepts you get a line of symmetry. \(\begin{align*} x =\frac {1+3}{2}=2 \end{align*} \)

y–intercept

Putting \(\begin{align*} x=0; \;\;y &= -(-1)(-3) =-3 \end{align*}\)


 

Example 1:

Given the quadratic function \(y = (x - 2)(x + 4)\)

  1. Find the coordinates of the x-intercepts and y-intercepts. 
  2. State the equation of the line of symmetry of the graph. 
  3. Find the coordinates of the turning point of the graph. State whether it is a maximum or minimum.
  4. Sketch the graph.

 

Solution:

The given equation is \(y = (x - 2)(x + 4)\)

A)  For x-intercepts:

     Upon substituting y = 0,

       \(\begin{align*} (x - 2)(x + 4) &= 0 \\ \\ ⇒\qquad\qquad (x - 2) &= 0\qquad \text{or}\qquad(x +4) = 0 \\ \\ ⇒\qquad\qquad\qquad\;\; x &= 2\qquad \text{or}\qquad       x = -4 \\ \\ \end{align*}\)

     Therefore, there are two x-intercepts obtained: \((2, 0)\) & \((-4, 0)\)

 

     For y-intercepts:

     Upon substituting \(x = 0\),

      \(\begin{align*} y &= (-2)(4) \\ \\ y &= -8 \end{align*} \)

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

 

B)  Taking the mean of the x-intercepts,

        \(\begin{align*} x &= \frac {2+(-4)}{2} \\ \\ x &= -22 \end{align*}\)

       ∴  \(\begin{align*} x= -1 \end{align*}\) is the equation of the line of symmetry of the graph.

 

C)  Since for the given equation, the coefficient of \(x^2 > 0\), the graph is an upward opening parabola and has a minimum turning point. 

     Putting \(x = -1\) in the original equation,

      \(\begin{align*} y &= (-1 - 2)(-1 + 4)\\ \\ y &= (-3)(3) \\ \\ y &= -9 \end{align*} \)

     Hence the turning point is at  \((-1, -9)\).

 

D)

S

Upward opening parabola

I

x-intercepts  \((2, 0 )\) & \((-4, 0 )\)

y-intercept \((0,8)\)

T

\((-1, -9 )\)

 

 

 

Conclusion

In this article we learnt how to determine the different parameters of the Quadratic function graphs and sketch graphs of Quadratic functions. 

 

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Further Trigonometry Quadratic Equations And Functions
Linear Inequalities Laws of Indices
Coordinate Geometry Graphs Of Functions And Graphical Solution
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