 4.05 Plotting parabolas

Lesson

Parabolas will always have one $y$y-intercept and can have zero, one or two $x$x-intercepts.

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 $y$y-intercept and any $x$x-intercepts.

Exploration

Find the $y$y-intercept and any $x$x-intercepts of the graph of $y=5x^2+25x-180$y=5x2+25x180.

First, let's factorise the right hand side of the equation.

 $y$y $=$= $5x^2+25x-180$5x2+25x−180 $y$y $=$= $5\left(x^2+5x-36\right)$5(x2+5x−36) Factorising the constant $5$5 from each term $y$y $=$= $5\left(x+9\right)\left(x-4\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 $y$y-intercept we set $x=0$x=0. This gives us $y=5\times9\times\left(-4\right)=-180$y=5×9×(4)=180. So the $y$y-intercept is at $\left(0,-180\right)$(0,180).

To find the $x$x-intercept we set $y=0$y=0. This gives us $0=5\left(x+9\right)\left(x-4\right)$0=5(x+9)(x4). Following the null factor law, the solutions are $x=-9$x=9 and $x=4$x=4. So there are two $x$x-intercepts, $\left(-9,0\right)$(9,0) and $4,0$4,0. The graph of $y=5x^2+25x-180$y=5x2+25x180 with the three intercepts marked.

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 $y$y-intercept by setting $x=0$x=0. This gives us $y=apq$y=apq, so the $y$y-intercept is $\left(0,apq\right)$(0,apq). And we can find the $x$x-intercepts 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 $x$x-intercepts are $\left(-p,0\right)$(p,0) and $\left(-q,0\right)$(q,0).

Summary

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:

• The $y$y-intercept will be $\left(0,apq\right)$(0,apq)
• The $x$x-intercepts will be $\left(-p,0\right)$(p,0) and $\left(-q,0\right)$(q,0)

We can use these three points to plot the parabola.

Practice questions

Question 1

Consider the equation $y=\left(1-x\right)\left(x+5\right)$y=(1x)(x+5).

1. State the $y$y-value of the $y$y-intercept.

2. Determine the $x$x-values of the $x$x-intercepts. Write all solutions on the same line separated by a comma.

3. Complete the table of values for the equation.

$x$x $-6$6 $-4$4 $-2$2 $0$0 $2$2
$y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
4. Determine the coordinates of the vertex of the parabola. State your answer in the form $\left(a,b\right)$(a,b).

5. Plot the graph of the parabola.

Question 2

Consider the equation $y=\left(x-1\right)\left(x-3\right)$y=(x1)(x3).

1. Is the parabola concave up or down?

Concave up

A

Concave down

B

Concave up

A

Concave down

B
2. Find the $x$x-values of the $x$x-intercepts of the curve.

3. Find the $y$y-value of the $y$y-intercept of the curve.

4. What is the equation of the vertical axis of symmetry for the graph?

5. Find the coordinates of the vertex of the parabola.

Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

6. Hence choose the correct curve for $y=\left(x-1\right)\left(x-3\right)$y=(x1)(x3).

A

B

C

D

A

B

C

D
Question 3

Consider the equation $y=2x+x^2$y=2x+x2.

1. Factorise the expression $2x+x^2$2x+x2.

2. Hence or otherwise solve for the $x$x-values of the $x$x-intercepts of $y=2x+x^2$y=2x+x2. Write all solutions on the same line separated by a comma.

3. Determine the $y$y-value of the $y$y-intercept of the graph.

4. Complete the table of values for the equation.

$x$x $-3$3 $-2$2 $-1$1 $0$0 $1$1
$y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
5. Determine the coordinates of the vertex of the parabola. State your answer in the form $\left(a,b\right)$(a,b).

6. Plot the graph of the parabola.