Study S3 Mathematics Quadratic Equations and Functions

Secondary 3

Maths

E-Maths

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 \(y = (x - 2)(x + 4)\),

Find the coordinates of the x-intercepts and y-intercepts.
State the equation of the line of symmetry of the graph.
Find the coordinates of the turning point of the graph. State whether it is a maximum or minimum.
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 \((-1, -9)\).

Upward opening parabola

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

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

\((-1, -9 )\)

Conclusion

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

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