Victorian Curriculum Year 10A - 2020 Edition
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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+25x180

 

$y$y $=$= $5\left(x^2+5x-36\right)$5(x2+5x36)

Factorising the constant $5$5 from each term

$y$y $=$= $5\left(x+9\right)\left(x-4\right)$5(x+9)(x4)

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.

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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).

    Loading Graph...

    A

    Loading Graph...

    B

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    C

    Loading Graph...

    D

    Loading Graph...

    A

    Loading Graph...

    B

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    C

    Loading Graph...

    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.

    Loading Graph...

Outcomes

VCMNA339

Explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital technology as appropriate

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