## Hwa Chong IP Sec 2 Maths Question – Equation of Parabola

Question:

Given that a parabola intersects the x-axis at x=-4 and x=2, and intersects the y-axis at y=-16, find the equation of the parabola.

Solution:

Sketch of graph: Now, there is a fast and slow method to this question. The slower method is to let $y=ax^2+bx+c$, and solve 3 simultaneous equations.

The faster method is to let $y=k(x+4)(x-2)$.

Why? We know that x=4 is a root of the polynomial, so it has a factor of (x-4). Similarly, the polynomial has a factor of (x-2). The constant k (to be determined) is added to scale the graph, so that the graph will satisfy y=-16 when x=0.

So, we just substitute in y=-16, x=0 into our new equation. $-16=k(4)(-2)$. $-16=-8k$.

So $k=2$.

In conclusion, the equation of the parabola is $y=2(x+4)(x-2)$. http://mathtuition88.com
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### 2 Responses to Hwa Chong IP Sec 2 Maths Question – Equation of Parabola

1. tomcircle says:

This one is similar, using “parameter k” technique on more challenging problem:

http://tomcircle.wordpress.com/2013/05/06/generalized-analytic-geometry/

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