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Stage 5.1-3

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 y-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 y-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

MA5.1-7NA

graphs simple non-linear relationships

MA5.2-10NA

connects algebraic and graphical representations of simple non-linear relationships

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