1. ## Turning 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?

2. 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:

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

So we set this equal to zero to get:

$\displaystyle 2x=0$ or x=0

3. 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:

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

So we set this equal to zero to get:

$\displaystyle 2x=0$ 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 $\displaystyle y = a(x - h)^2 + k$ has a turning point at (h, k).

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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