2.06 Quadratic functions

Key features

We've already looked at how to find certain features of linear equations including the gradient, the $x$x-intercept, and the $y$y-intercept. Now we are going to explore the features of parabolas.

Intercepts

You can see them in the picture below. 

Recall that we may have none, one or two $x$x-intercepts depending on the number of solutions to the quadratic equation where $y=0$y=0.

Maximum and minimum values

Maximum or minimum values are also known as the turning points, and they are found at the vertex of the parabola.

Parabolas that are concave up have a minimum value. This means the $y$y-value will never go under a certain value.	Parabolas that are concave down have a maximum value. This means the $y$y-value will never go over a certain value.

Axis of symmetry

Maximum and minimum values occur on a parabola's axis of symmetry. This is the vertical line that evenly divides a parabola into two sides down the middle. Using the general form of the quadratic $y=ax^2+bx+c$y=ax2+bx+c, substitute the relevant values of $a$a and $b$b into:

Does it look familiar? This is part of the quadratic formula. It is the half-way point between the two solutions. Note that the vertex always sits on the axis of symmetry so when we find the equation of the axis of symmetry we are also finding the $x$x value of the vertex. 

Gradient of a curve

We have learned about positive and negative gradients when we looked at straight lines. Generally, straight lines had only one or the other.

However, a parabola has a positive gradient in some places and negative gradient in others. The parabola's maximum or minimum value, at which the gradient is $0$0, is the division between the positive and negative gradient regions. Hence why it is also called the turning point.

Look at the picture below. See how one side of the parabola has a positive gradient, then there is a turning point with a zero gradient, then the other side of the parabola has a negative gradient?

We can see in this particular graph that the gradient is positive when $x<1$x<1 and we can say that the parabola is increasing for these values of $x$x. Similarly, the gradient is negative when $x>1$x>1, and for these values of $x$x the parabola is decreasing.

Practice questions

Question 1

Question 2

Graphing quadratics

We can be expected to graph quadratic functions from any of the following forms:

Each form has an advantage for different key features that can be identified quickly. However, for each form the role of $a$a remains the same.

That is:

• If $a<0$a<0 then the quadratic is concave down
• If $a>0$a>0 then the quadratic is concave up
• The larger the magnitude of $a$a, the narrower the curve

Let's now look at examples for how to find all the key features for each form.

Graphing from turning point form

$y=a\left(x-h\right)^2+k$y=a(x−h)2+k

• This form is given this name as we can read the turning point directly from the equation: $(h,k)$(h,k)
• We can obtain this graph from the graph of $y=x^2$y=x2, by dilating(stretching) the graph by a factor of $a$a from the $x$x-axis, then translating the graph $h$h units horizontally and $k$k units vertically
• The axis of symmetry will be at $x=h$x=h
• The $y$y-intercept can be found by substituting $x=0$x=0 into the equation
• The $x$x-intercept(s) can be found by substituting $y=0$y=0 into the equation and solving

Practice questions

Question 3

Question 4

Graphing from factored form

$y=a\left(x-\alpha\right)\left(x-\beta\right)$y=a(x−α)(x−β)

• From this form we can read the $x$x-intercepts directly: $(\alpha,0)$(α,0) and $(\beta,0)$(β,0). Since, if $y=0$y=0 we can solve the equation using the null factor law.
• The axis of symmetry will be at the $x$x-coordinate midway between the two $x$x-intercepts. Alternatively you could expand this equation to general form and then use the formula for the axis of symmetry: $x=\frac{-b}{2a}$x=−b2a​.
• The axis of symmetry is also the $x$x-coordinate of the turning point. Substitute this value into the equation the find the $y$y-coordinate of the turning point.
• The $y$y-intercept can be found by substituting $x=0$x=0 into the equation

Practice question

Question 5

Graphing from general form

$y=ax^2+bx+c$y=ax2+bx+c

• From this form we can read the $y$y-intercept directly: $(0,c)$(0,c) 
• Find the axis of symmetry using the formula: $x=\frac{-b}{2a}$x=−b2a​
• The axis of symmetry is also the $x$x-coordinate of the turning point. Substitute this value into the equation the find the $y$y-coordinate of the turning point.
• The $x$x-intercept can be found by substituting $y=0$y=0 into the equation and then using one of our methods for solving quadratics, such as factorising or the quadratic formula, to find the zeros of the equation.

Alternatively, we can use the method of completing the square to rewrite the quadratic in turning point form.

Practice questions

Question 6

question 7

Graphing quadratics using technology

Technology can be used to graph and find key features of a quadratic function. We may need to use our knowledge of the graph's domain and range or calculate the location of some key features to find an appropriate view window.

Practice question

Question 8