4.05 Graphing quadratic functions
Now that we have observed some of the key characteristics of quadratic functions, let's look at how we might create a graph of a quadratic function.
Graphing from a table of values
Recall that you can create a graph of a function by generating a table of values and evaluating the function for certain values in its domain. We can do the same for quadratic functions, and connect the points in a smooth curve that looks like a parabola.
Practice questions
Question 1
Consider the function $y=x^2$y=x2
Complete the following table of values.
$x$x $-3$−3 $-2$−2 $-1$−1 $0$0 $1$1 $2$2 $3$3 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the points in the table of values.
Loading Graph...Hence plot the curve.
Loading Graph...Are the $y$y values ever negative?
No
AYes
BWrite down the equation of the axis of symmetry.
What is the minimum $y$y value?
For every $y$y value greater than $0$0, how many corresponding $x$x values are there?
$3$3
A$1$1
B$2$2
C
Question 2
Consider the function $y=\left(x-2\right)^2$y=(x−2)2
Complete the following table of values.
$x$x $0$0 $1$1 $2$2 $3$3 $4$4 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Sketch a graph of the function.
Loading Graph...What is the minimum $y$y value?
What $x$x value corresponds to this minimum $y$y value?
What are the coordinates of the vertex? Give your answer in the form $\left(a,b\right)$(a,b).
Graphing from characteristics in standard form
It's also possible to sketch a graph of a parabola given certain key characteristics. We can determine these key characteristics more easily by writing the quadratic equation in standard form.
The standard form of a quadratic equation is $y=ax^2+bx+c$y=ax2+bx+c, where $a\ne0$a≠0.
Given a quadratic equation in standard form, we can identify certain key characteristics and use them to graph the function.
Key characteristics
- Line of symmetry (axis of symmetry): Defined by the equation $x=\frac{-b}{2a}$x=−b2a, which is a vertical line through the vertex
- Vertex (turning point): Can be found using technology or by substituting in the line of symmetry for the $x$x-value to find the $y$y-value
- $y$y-intercept: Can be found by substituting in $x=0$x=0 and evaluating
- Direction of opening: The parabola will open up if $a>0$a>0 and will open down if $a<0$a<0
Using these key characteristics, we can graph the parabola. For a better shape, we can substitute in a few $x$x-values to get a few more points.
Worked example
Question 3
Graph the quadratic function $y=x^2-6x+4$y=x2−6x+4
Think: We need to identify the key characteristics and then we can graph the parabola.
Do:
1. Find the line of symmetry using $x=\frac{-b}{2a}$x=−b2a where $b=-6$b=−6 and $a=1$a=1
| $x$x | $=$= | $\frac{-b}{2a}$−b2a |
| $x$x | $=$= | $\frac{-\left(-6\right)}{2\times1}$−(−6)2×1 |
| $x$x | $=$= | $\frac{6}{2}$62 |
| $x$x | $=$= | $3$3 |
2. Find the y-value of the vertex by substituting in $x=3$x=3.
| $y$y | $=$= | $x^2-6x+4$x2−6x+4 |
| $y$y | $=$= | $3^2-6\times3+4$32−6×3+4 |
| $y$y | $=$= | $9-18+4$9−18+4 |
| $y$y | $=$= | $-5$−5 |
The vertex is $\left(3,-5\right)$(3,−5) with $a=1$a=1, so the parabola will open up making $\left(3,-5\right)$(3,−5) a minimum.
3. Find the $y$y-intercept, but substituting in $x=0$x=0.
| $y$y | $=$= | $x^2-6x+4$x2−6x+4 |
| $y$y | $=$= | $0^2-6\times0+4$02−6×0+4 |
| $y$y | $=$= | $0-0+4$0−0+4 |
| $y$y | $=$= | $4$4 |
So the y-intercept is $\left(0,4\right)$(0,4).
Now we put it all together on a graph.
Practice questions
Question 4
Consider the curve $y=x^2+6x+4$y=x2+6x+4.
Determine the axis of symmetry.
Hence determine the minimum value of $y$y.
Using the minimum point on the curve, plot the graph of the function.
Loading Graph...
Question 5
Consider the quadratic $y=3x^2-6x+8$y=3x2−6x+8
Find the axis of symmetry.
Find the vertex of the parabola. Give your answer in the form $\left(a,b\right)$(a,b).
What is the $y$y value of the $y$y-intercept of this quadratic function?
Plot the function.
Loading Graph...