Derivation for finding vertex and turning point of parabola? My teacher told me that form the equation of parabola .
$y + \frac{D}{4a} = a(x+\frac{b}{2a})^2$
So , to find vertex of turning point ,
Make $y=0$ and $x = 0$.
You get $-\frac{b}{2a}$ and $-\frac{D}{4a}$ but aren’t we getting these values in +?
I am not getting why would he say that and how to reach there.
Please do share if there is any other point you have.
 A: The "a" in the vertex form is the same "a" as
in $y = ax2 + bx + c$ (that is, both a's have exactly the same value). The sign on "a" tells you whether the quadratic opens up or opens down. Think of it this way: A positive "a" draws a smiley, and a negative "a" draws a frowny. (Yes, it's a silly picture to have in your head, but it makes is very easy to remember how the leading coefficient works.)
In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.
Suppose that "a" is positive, so a(x – h)2 is zero or positive and, whatever x-value you choose, you're always taking k and adding a(x – h)2 to it. That is, the smallest value y can be is just k; otherwise y will equal k plus something positive. When does y equal only k? When x – h, the squared part, is zero; in other words, when x = h. So the lowest value that y can have, y = k, will only happen if x = h. And the lowest point on a positive quadratic is of course the vertex.
If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y can achieve is y = k at x = h. And the highest point on a negative quadratic is of course the vertex.
A: It is easier to understand it if we substitute variables.
Let:
$$p = y + \frac{D}{4a}$$
$$s = x+\frac{b}{2a}$$
Then the equation transforms to
$$p = as^2$$
Where is the vertex of this parabola? Obviously at $p = 0, s = 0$.
Vertex at $p = 0$ means it is at $y + \frac{D}{4a} = 0$, which means $y = -\frac{D}{4a}$.
Vertex at $s = 0$ means it is at $x+\frac{b}{2a} = 0$, which means $x = -\frac{b}{2a}$.
A: To find the vertex of a parabola, equate the slope to 0.
that is $dy/dx=0$
$dy/dx= 2a(x+\frac{b}{2a})$
so
$2a(x+\frac{b}{2a})=0$
so $x=-b/2a$
putting the point in the equation, we get the y coordinates as $-\frac{D}{4a}$
