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?
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?
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).
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.