Parabolas will always have one $y$yintercept and can have zero, one or two $x$xintercepts.
Quadratic expressions can be factorised, so that $ax^2+a\left(p+q\right)x+apq=a\left(x+p\right)\left(x+q\right)$ax2+a(p+q)x+apq=a(x+p)(x+q).
Combining these two ideas, we can use factorisation to plot parabolas. If a quadratic equation can be factorised into the form $y=a\left(x+p\right)\left(x+q\right)$y=a(x+p)(x+q) then we can immediately find the $y$yintercept and any $x$xintercepts.
Find the $y$yintercept and any $x$xintercepts of the graph of $y=5x^2+25x180$y=5x2+25x−180.
First, let's factorise the right hand side of the equation.
$y$y  $=$=  $5x^2+25x180$5x2+25x−180 

$y$y  $=$=  $5\left(x^2+5x36\right)$5(x2+5x−36) 
Factorising the constant $5$5 from each term 
$y$y  $=$=  $5\left(x+9\right)\left(x4\right)$5(x+9)(x−4) 
Using the fact that $9+\left(4\right)=5$9+(−4)=5 and $9\times\left(4\right)=36$9×(−4)=−36 
To find the $y$yintercept we set $x=0$x=0. This gives us $y=5\times9\times\left(4\right)=180$y=5×9×(−4)=−180. So the $y$yintercept is at $\left(0,180\right)$(0,−180).
To find the $x$xintercept we set $y=0$y=0. This gives us $0=5\left(x+9\right)\left(x4\right)$0=5(x+9)(x−4). Following the null factor law, the solutions are $x=9$x=−9 and $x=4$x=4. So there are two $x$xintercepts, $\left(9,0\right)$(−9,0) and $4,0$4,0.
We can generalise from this. If we have a quadratic equation of the form $y=a\left(x+p\right)\left(x+q\right)$y=a(x+p)(x+q), then we can find the $y$yintercept by setting $x=0$x=0. This gives us $y=apq$y=apq, so the $y$yintercept is $\left(0,apq\right)$(0,apq). And we can find the $x$xintercepts by setting $y=0$y=0. This gives us $0=a\left(x+p\right)\left(x+q\right)$0=a(x+p)(x+q), so that either $x=p$x=−p or $x=q$x=−q, and the two $x$xintercepts are $\left(p,0\right)$(−p,0) and $\left(q,0\right)$(−q,0).
If we can factorise the equation of a parabola into the form $y=a\left(x+p\right)\left(x+q\right)$y=a(x+p)(x+q) then:
We can use these three points to plot the parabola.
Consider the equation $y=\left(1x\right)\left(x+5\right)$y=(1−x)(x+5).
State the $y$yvalue of the $y$yintercept.
Determine the $x$xvalues of the $x$xintercepts. Write all solutions on the same line separated by a comma.
Complete the table of values for the equation.
$x$x  $6$−6  $4$−4  $2$−2  $0$0  $2$2 

$y$y  $\editable{}$  $\editable{}$  $\editable{}$  $\editable{}$  $\editable{}$ 
Determine the coordinates of the vertex of the parabola. State your answer in the form $\left(a,b\right)$(a,b).
Plot the graph of the parabola.
Consider the equation $y=\left(x1\right)\left(x3\right)$y=(x−1)(x−3).
Is the parabola concave up or down?
Concave up
Concave down
Concave up
Concave down
Find the $x$xvalues of the $x$xintercepts of the curve.
Find the $y$yvalue of the $y$yintercept of the curve.
What is the equation of the vertical axis of symmetry for the graph?
Find the coordinates of the vertex of the parabola.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Hence choose the correct curve for $y=\left(x1\right)\left(x3\right)$y=(x−1)(x−3).
Consider the equation $y=2x+x^2$y=2x+x2.
Factorise the expression $2x+x^2$2x+x2.
Hence or otherwise solve for the $x$xvalues of the $x$xintercepts of $y=2x+x^2$y=2x+x2. Write all solutions on the same line separated by a comma.
Determine the $y$yvalue of the $y$yintercept of the graph.
Complete the table of values for the equation.
$x$x  $3$−3  $2$−2  $1$−1  $0$0  $1$1 

$y$y  $\editable{}$  $\editable{}$  $\editable{}$  $\editable{}$  $\editable{}$ 
Determine the coordinates of the vertex of the parabola. State your answer in the form $\left(a,b\right)$(a,b).
Plot the graph of the parabola.
Explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital technology as appropriate