I don't know if this thread is right for the forum..

but I have a maths assignment where it's asking to state the turning point of certain equations but i don't know how to find it.

e.g

y=x^2

how to i find the turning point of that parabola?

Printable View

- Aug 24th 2009, 02:14 AMamberley-janeTurning Points
I don't know if this thread is right for the forum..

but I have a maths assignment where it's asking to state the turning point of certain equations but i don't know how to find it.

e.g

y=x^2

how to i find the turning point of that parabola? - Aug 24th 2009, 02:27 AMKiwi_Dave
At the turning points of an equation the slope of y is zero. In other words the differential of the equation must be zero. So for your example:

So we set this equal to zero to get:

or x=0 - Aug 24th 2009, 02:42 AMmr fantastic
Since the question is posted in the Pre-calculus subforum I have to assume that a non-calculus approach is required.

@OP: You're probably expected to know the turning point form of a parabola and how to get it. Certainly you should know that has a turning point at (h, k). - Aug 24th 2009, 03:30 AMProve It
There are three approaches to finding the turning point of a parabola.

Method 1) Due to the symmetry of the parabola, the turning point lies halfway between the x-intercepts.

If there is only one x-intercept, then the x-intercept IS the turning point.

So in your case, for , the x-intercept is found by letting .

So .

Since there is only one x-intercept, that is the x-coordinate of the turning point.

To find the y-coordinate, substitute it into your equation.

.

So the turning point is .

For another example, look at the parabola given by the equation .

To find the x-intercept, let .

So

or

or .

To find the x-coordinate of the turning point, average the x-intercepts.

So .

To find the y-coordinate, substitute this value into the equation.

.

So the turning point is .

Method 2)

Method 1 is ok to use if x-intercepts exist. However, it is inappropriate to use this technique if x-intercepts do not exist.

Therefore, most people prefer to use the "Turning Point Form" of the parabola.

For any parabola written in the form

its turning point is at .

This form of the parabola is found by completing the square.

With your example, , we can rewrite it as

.

So its turning point is .

With the other example, , we complete the square to put it into turning point form.

.

So the turning point is .

Method 3) is to use Differential Calculus, but since this is in the Pre-Calculus forum, I doubt you'd be using this method.