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Year 10AC9M10A05

The Parabola — Given the Turning Point

Experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns.

Things you should know.

Method

Standard form
y=ax2+qy = ax^{2} + q
$q$ is the $y$ intercept, and the turning point sits on the $y$ axis.
Intercept form
y=a(x−p)(x−q)y = a(x - p)(x - q)
$p$ and $q$ are the $x$ intercepts.
Shape
a>0 opens upa<0 opens downa > 0 \ \text{opens up} \qquad a < 0 \ \text{opens down}
The sign of $a$ decides which way it faces.

Worked example

e.g. given (0;4) and (1;10)\text{given } (0;4) \text{ and } (1;10)
  1. Write the equation in the form y=ax2+qy = ax^{2} + qy=ax2+qy = ax^{2} + q
  2. Substitute the qq value. It is the yy intercept, so q=4q = 4y=ax2+4y = ax^{2} + 4
  3. Substitute the other coordinates into xx and yy10=a(1)2+410 = a(1)^{2} + 4
  4. Calculate aaa=6a = 6
  5. Substitute back into step 2y=6x2+4y = 6x^{2} + 4
    The parabola y equals six x squared plus four through both given pointsxyy = 6x² + 4(0;4)(1;10)

Practice

Determine the equation of the parabola passing through:

1
(0;3) and (1;5)(0;3) \text{ and } (1;5)
Answery=2x2+3y = 2x^{2} + 3
2
(0;−1) and (2;7)(0;-1) \text{ and } (2;7)
Answery=2x2−1y = 2x^{2} - 1
3
(0;5) and (1;2)(0;5) \text{ and } (1;2)
Answery=−3x2+5y = -3x^{2} + 5
4
(0;0) and (2;8)(0;0) \text{ and } (2;8)
Answery=2x2y = 2x^{2}
5
(0;6) and (3;−3)(0;6) \text{ and } (3;-3)
Answery=−x2+6y = -x^{2} + 6
6
(0;−4) and (1;−1)(0;-4) \text{ and } (1;-1)
Answery=3x2−4y = 3x^{2} - 4
Next step
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