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Year 10AC9M10A05

The Parabola — Given the Intercepts

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

Worked example

e.g. given (0;2), (2;0) and (2;0)\text{given } (0;-2),\ (-2;0) \text{ and } (2;0)
  1. Write the equation in the form y=a(xp)(xq)y = a(x - p)(x - q)y=a(xp)(xq)y = a(x - p)(x - q)
  2. Substitute the two xx intercepts. Watch the signsy=a(x2)(x+2)y = a(x - 2)(x + 2)
  3. Substitute the other coordinate to calculate aa2=a(02)(0+2)  a=12-2 = a(0-2)(0+2) \ \therefore \ a = \tfrac{1}{2}
  4. Substitute aa back in and expand to the form y=ax2+qy = ax^{2} + qy=12x22y = \tfrac{1}{2}x^{2} - 2
    A parabola crossing the x axis at negative two and twoxyy = ½x² - 2(0;-2)

Practice

Determine the equation of the parabola with:

1
x intercepts 1,3 and passing through (0;3)x \text{ intercepts } -1, 3 \text{ and passing through } (0;-3)
Answery=x22x3y = x^{2} - 2x - 3
2
x intercepts 3,3 and passing through (0;9)x \text{ intercepts } -3, 3 \text{ and passing through } (0;-9)
Answery=x29y = x^{2} - 9
3
x intercepts 0,4 and passing through (2;4)x \text{ intercepts } 0, 4 \text{ and passing through } (2;-4)
Answery=x24xy = x^{2} - 4x
4
x intercepts 1,1 and passing through (0;3)x \text{ intercepts } -1, 1 \text{ and passing through } (0;3)
Answery=3x2+3y = -3x^{2} + 3
5
x intercepts 2,5 and passing through (0;10)x \text{ intercepts } 2, 5 \text{ and passing through } (0;10)
Answery=x27x+10y = x^{2} - 7x + 10
6
x intercepts 4,2 and passing through (0;8)x \text{ intercepts } -4, 2 \text{ and passing through } (0;-8)
Answery=x2+2x8y = x^{2} + 2x - 8
Next step
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