Topics
4 Quadratics and polynomials
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Australia
10&10a

4.05 Plotting parabolas

Lesson
Worksheet
Practice
Lesson

Plot the parabolas

 Parabolas  will always have one y-intercept and can have zero, one or two 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).

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) then we can immediately find the y-intercept and any x-intercepts.

If we have a quadratic equation of the form y=a(x+p)(x+q), then we can find the y-intercept by setting x = 0. This gives us y = apq, so the y-intercept is \left( 0, apq \right). And we can find the y-intercepts by setting y = 0. This gives us 0 = a \left( x + p \right) \left( x + q \right), so that either x = -p or x = -q, and the two x-intercepts are \left( -p,\, 0 \right) and \left( -q,\, 0 \right).

If we can factorise the equation of a parabola into the form y = a \left( x + p \right) \left( x + q \right) then:

  • The y-intercept will be \left( 0,\, apq \right)

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

We can use these three points to plot the parabola.

Examples

Example 1

Consider the equation y = \left(2 - x\right) \left(x + 4\right).

a

State the y-value of the y-intercept.

Worked Solution
Create a strategy

Substitute x=0 into the equation and then solve for y.

Apply the idea
\displaystyle y\displaystyle =\displaystyle (2-x)(x+4)Write the equation
\displaystyle y\displaystyle =\displaystyle \left(2 - 0\right) \left(0 + 4\right)Substitute x=0
\displaystyle =\displaystyle 8Evaluate the product
b

Determine the x-values of the x-intercepts.

Worked Solution
Create a strategy

Substitute y=0 into the equation and then solve for x.

Apply the idea
\displaystyle y\displaystyle =\displaystyle \left( 2-x \right) \left( x+4 \right)Write the equation
\displaystyle 0\displaystyle =\displaystyle \left(2 - x\right) \left(x + 4\right)Substitute y=0
\displaystyle 2-x\displaystyle =\displaystyle 0Equate the first factor to 0
\displaystyle x\displaystyle =\displaystyle 2Add x to both sides
\displaystyle x+4\displaystyle =\displaystyle 0Equate the second factor to 0
\displaystyle x\displaystyle =\displaystyle -4Subtract 4 from both sides

So the x-intercepts are x=2,\,x=-4.

c

Determine the coordinates of the vertex of the parabola.

Worked Solution
Create a strategy

The x-coordinate of the vertex is half-way between the x-intercepts.

Apply the idea
\displaystyle x\displaystyle =\displaystyle \dfrac{2+(-4)}{2}Find the average of the x-intercepts
\displaystyle =\displaystyle -1Evaluate
\displaystyle y\displaystyle =\displaystyle (2-(-1))(-1+4)Substitute x=-1 into the equation
\displaystyle =\displaystyle 9Evaluate

The coordinates of the vertex are (-1,9).

d

Plot the graph of the parabola.

Worked Solution
Create a strategy

Plot the intercepts and the vertex and draw a curve through the points.

Apply the idea
-5
-4
-3
-2
-1
1
2
3
x
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
y

Example 2

Consider the equation y=4x+x^{2}.

a

Factorise the expression 4x+x^{2}.

Worked Solution
Create a strategy

Take out the common factor of the two terms.

Apply the idea

The common factor of the two terms is x.

4x+x^{2}=x(4+x)

b

Solve for the x-values of the x-intercepts of y=2x+x^{2}.

Worked Solution
Create a strategy

Substitute y=0 into the equation and then solve for x.

Apply the idea
\displaystyle y\displaystyle =\displaystyle 4x+x^{2}Write the equation
\displaystyle y\displaystyle =\displaystyle x \left(4 + x\right)Factorise the quadratic equation
\displaystyle 0\displaystyle =\displaystyle x \left(4 + x\right)Substitute y=0
\displaystyle x\displaystyle =\displaystyle 0Equate the first factor to 0
\displaystyle 4 + x\displaystyle =\displaystyle 0Equate the second factor to 0
\displaystyle x\displaystyle =\displaystyle -4Subtract 4 from both sides

So the x-intercepts are x=0,\,x=-4.

c

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

Worked Solution
Create a strategy

Substitute x=0 into the equation then solve for y.

Apply the idea
\displaystyle y\displaystyle =\displaystyle 4x+x^{2}Write the equation
\displaystyle =\displaystyle 4 \times 0 + 0^{2}Substitute x=0
\displaystyle =\displaystyle 0Evaluate
d

Find the coordinates of the vertex.

Worked Solution
Create a strategy

Use the equation x=-\dfrac{b}{2a} and substitute the value of x into the equation of the parabola.

Apply the idea

From the equation, y=4x+x^2, \, a=1,\,b=4.

\displaystyle x\displaystyle =\displaystyle -\dfrac{4}{2\times 1}Substitute a and b
\displaystyle =\displaystyle -2Evaluate

The equation of the axis of symmetry is x=-2, which is also the x-coordinate of the vertex.

\displaystyle y\displaystyle =\displaystyle 4x+x^2Write the equation
\displaystyle =\displaystyle 4(-2)+(-2)^2Substitute x=-2
\displaystyle =\displaystyle -4Evaluate

The coordinates of the vertex is (-2,-4).

e

Plot the graph of the parabola.

Worked Solution
Create a strategy

Plot the intercepts and vertex and draw a curve through the points.

Apply the idea
-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y
Idea summary

If we can factorise the equation of a parabola into the form y = a \left( x + p \right) \left( x + q \right) then:

  • The y-intercept will be \left( 0,\, apq \right)

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

We can use these three points to plot the parabola.

Outcomes

ACMNA239

Explore the connection between algebraic and graphical representations of relations such as simple quadratics, circles and exponentials using digital technology as appropriate

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