21.05 Quadratic functions

Key features

Previously, key features of linear equations have been explored including the gradient, the $x$x-intercept, and the $y$y-intercept. The graph of a quadratic equation is a parabola and is a curved or concave shape (either concave up or down, depending on the equation). The following sections describe the key features of a parabola.

 

Intercepts

These intercepts are shown in the picture below: 

Recall that there may be one or two or even no $x$x-intercepts:

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 the following equation:

Does it look familiar? This is part of the quadratic formula. It is the half-way point between the two solutions.

 

The gradient of a curve

Positive and negative gradients were previously explored for straight lines. If we draw the tangent line to a curve at a particular point we can estimate the gradient of a curve. 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. One side of the parabola has a positive gradient, there is a turning point with a zero gradient, and the other side of the parabola has a negative gradient.

In this particular graph, the gradient is positive when $x<1$x<1 and 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.

Worked example

example 1

The above graph shows the quadratic function $y=x^2+3x-4.$y=x2+3x−4. 

a) Use the graph to solve the equation $x^2+3x-4=0.$x2+3x−4=0.

b) Use the graph to approximate the solutions to the equation $x^2+3x-4=2$x2+3x−4=2.

a)

Think: The solutions of the equation will equal the $x$x-coordinates of the $x$x-intercepts of the parabola, since $y=0$y=0 at these points. 

Do: We can see that the $x$x-intercepts occur at $x=-1$x=−1 and $4$4. Therefore the solutions to the equation $x^2+3x-4=0$x2+3x−4=0 are$x=-1,4$x=−1,4.

b)

Think: To find the solutions to this equation we should approximate the $x$x-coordinates of the points of intersection of the parabola with the horizontal line $y=2$y=2. 

Do: By looking at where the parabola intersects the horizontal grid line at $y=2$y=2, we can approximate the coordinates of these points to be $(-0.5,2)$(−0.5,2) and $(3.5,2)$(3.5,2).

Therefore estimates to the solutions of the equation $x^2+3x-4=2$x2+3x−4=2 are $x\approx-0.5,3.5$x≈−0.5,3.5.

 

Practice questions

Question 1

Question 2

Graphing quadratics

As for linear equations, there are a number of forms for graphing quadratic functions:

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

 

The following are examples of 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 the turning point can be read directly from the equation: $(h,k)$(h,k)
• 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 can be found by substituting $y=0$y=0 into the equation and rearranging

 

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, read the $x$x-intercepts directly: $(\alpha,0)$(α,0) and $(\beta,0)$(β,0). If  $y=0$y=0, 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, 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 then 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, 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, then 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 the methods for solving quadratics, such as factorising or the quadratic formula, to find the zeros (x - intercepts) of the equation.

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

Sometimes, the solutions for a quadratic may be asked for. This is referring to the $x$x-intercepts which satisfy the equation $y=0$y=0. The solutions are sometimes called the roots of the equation.

 

Practice questions

Question 6

question 7

Question 8

 

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 or calculate the location of some key features to find an appropriate view window.

Practice question

Question 9

 

Let's explore how a quadratic function changes using the applet below. What happens when we change the sliders for dilation, reflection, vertical translation and horizontal translation? Can you describe what changes these values make to the quadratic function?

These movements are called transformations. Transform means change, and these transformations change the simple quadratic $y=x^2$y=x2 into other quadratics by moving (translating), flipping (reflecting) and making the graph appear more or less steep (dilating).

By using the above applet, step through these instructions:

• Start with the simple quadratic $y=x^2$y=x2
• Dilate the quadratic by factor of $2$2
• Reflect on the $x$x-axis
• Translate vertically by $2$2 and horizontally by $-3$−3 units

How has the graph changed? Can you visualise the changes without using the applet? What is the resulting equation?

 

Equations and Transformations of Quadratics

Transforming a quadratic will change its equation. Here are some of the most common types of quadratic equations, and what they mean with regards to the transformations that have occurred.

$y=ax^2$y=ax2

If $a<0$a<0 (that is, $a$a is negative) then we have a reflection parallel to the $x$x-axis. It's like the quadratic has been flipped upside down.

 

$y=ax^2$y=ax2

This is a quadratic that has been dilated vertically by a factor of $a$a.

If $\left|a\right|>1$|a|>1 then the graph is steeper than $y=x^2$y=x2.

If $\left|a\right|<1$|a|<1 then the graph is flatter than $y=x^2$y=x2. 

 

$y=ax^2+k$y=ax2+k

In the graph $y=ax^2+k$y=ax2+k, the quadratic has been vertically translated by $k$k units.  

If $k>0$k>0 then the translation is up.

If $k<0$k<0 then the translation is down.

 

$y=\left(x-h\right)^2$y=(x−h)2

The $h$h indicates the horizontal translation.  

If $h>0$h>0, that is the factor in the brackets is $\left(x-h\right)$(x−h) than we have a horizontal translation of $h$h units right.

If $h<0$h<0, that is the factor in the brackets is $\left(x-\left(-h\right)\right)=\left(x+h\right)$(x−(−h))=(x+h) than we have a horizontal translations of $h$h units left.  

 

 

Identifying transformations

The form $y=a\left(x-h\right)^2+k$y=a(x−h)2+k shows us:

Vertical Dilation a

If $\left|a\right|>1$|a|>1 then the quadratic is steeper than $x^2$x2.

If $\left|a\right|<1$|a|<1 then the quadratic is flatter than $x^2$x2.

Reflection a

If $a<0$a<0 then the quadratic is a reflection of $ax^2$ax2 on a line parallel to the $x$x-axis.

Vertical Translation k

If $k>0$k>0 then the translation is up.

If $k<0$k<0 then the translation is down.

Horizontal Translation h

If $h<0$h<0 then the translation is to the left, and $x+h$x+h appears in the brackets.

If $h>0$h>0 then the translation is to the right, and $x-h$x−h appears in the brackets.

 

Practice questions

Question 10

Question 11

Question 12

Question 13