Someone please help me, who is good in math?

[j_sun]

My homework is killing me.

What is the vertex of this function.
f(x)= (2p) x ^2 - (12p)x + (3p)

[hummingbird]

check this for a typo in the last term

[Melanie]

f(x) = (2p)x² - (12p)x + (3p)

First we need the axis of symmetry which is -b/2a with quadratics in the form of ax² + bx + c = 0

In your case, a = 2p and b = -12p

Therefore, your axis of symmetry is:

12p / 4p = 3

x = 3

Plug this x-value back into the equation:

f(3) = (2p)*3² - (12p)*3 + (3p)
f(3) = (18p) - (36p) + (3p)
f(3) = -15p

===================================

Your vertex is:

(3, -15p)

[camila]

The x-coordinate of the vertex is given by the formula
x = -b/2a
so in this case,
x = -(-12p)/(2*2p)
=12p/4p
x = 3

The y-coordinate can be found by finding what y is equal to for that value of x:
f(3) = 2p(3^2) - (12p)*3 + 3p
=18p - 36p + 3p
=-15p

So the vertex is at the point (3, -15p).

[camila]

It is a quadratic where the coefficients are factors of p instead of being "normal" numbers.

The vertex is a critical point: a maximum or minimum.  It is where the slope is equal to zero.

To find the slope, you can differentiate:

f(x) = 2p x^2 - 12p x + 3p
f'(x) = 2(2p)x - 12p
f'(x) = 4p(x) - 12p
f'(x) = 4p (x - 3)

Since p cannot be zero (otherwise your graph would be a flat line -- the x axis), then (x-3) must be zero, and that will only happen when x = 3

To find the vertex point, set x = 3 in the original equation:
f(x) = 2p x^2 - 12p x + 3p
f(3) = 2p(9) -12p(3) + 3p
f(3p) = 18p -36p + 3p = -15p


Vertex is at (x, f(x)) = (3, -15p)

[federox]

f(x) = 2px² - 12px + 3p
f(x) = 2p(x² - 6x ....) + 3p
f(x) = 2p(x² - 6x + 9) + 3p - 18p
f(x) = 2p(x - 3)² - 15p
vertex is (3, -15p)