Turning Points

**amberley-jane** · August 24th 2009, 01:14 AM

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?

**Kiwi_Dave** · August 24th 2009, 01:27 AM

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:

$\frac {dy}{dx}=2x$

So we set this equal to zero to get:

$2x=0$ or x=0

**mr fantastic** · August 24th 2009, 01:42 AM

Originally Posted by Kiwi_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:

$\frac {dy}{dx}=2x$

So we set this equal to zero to get:

$2x=0$ or x=0

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 $y = a(x - h)^2 + k$ has a turning point at (h, k).

**Prove It** · August 24th 2009, 02:30 AM

Originally Posted by amberley-jane

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?

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 $y = x^2$ , the x-intercept is found by letting $y = 0$ .

So $0 = x^2 \implies x = 0$ .

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.

$y = 0^2 \implies y = 0$ .

So the turning point is $(x, y) = (0, 0)$ .

For another example, look at the parabola given by the equation $y = x^2 + 3x + 2$ .

To find the x-intercept, let $y = 0$ .

So $0 = x^2 + 3x + 2$

$0 = (x + 1)(x + 2)$

$x + 1 = 0$ or $x + 2 = 0$

$x = -1$ or $x = -2$ .

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

So $x_{tp} = \frac{(-1) + (-2)}{2} = -\frac{3}{2}$ .

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

$y_{tp} = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) + 2$

$= \frac{9}{4} - \frac{9}{2} + 2$

$= -\frac{1}{4}$ .

So the turning point is $(x, y) = \left(-\frac{3}{2}, -\frac{1}{4}\right)$ .

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

$y = a(x - h)^2 + k$

its turning point is at $(x, y) = (h, k)$ .

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

With your example, $y = x^2$ , we can rewrite it as

$y = (x - 0)^2 + 0$ .

So its turning point is $(x, y) = (0, 0)$ .

With the other example, $y = x^2 + 3x + 2$ , we complete the square to put it into turning point form.

$y = x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 2$

$= \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + 2$

$= \left[x - \left(-\frac{3}{2}\right)\right]^2 - \frac{1}{4}$ .

So the turning point is $(x, y) = \left( -\frac{3}{2}, \, -\frac{1}{4}\right)$ .

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

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