Help with finding a vertex I am having trouble with this math problem.
The coordinate of the vertex of $f(x) = ax^2 + bx +c, a \ne 0$, is______.
Now I know that to find the $x$ coordinate I would use this formula.
$x = \frac{-b}{2a}$
However I am not sure how to find the y coordinate. I also don't know how to find the $x$ coordinate without numbers for $b$ and $a$. I assume that it would just be $\frac{-b}{2a}$ but I am not sure. Any help would be appreciated, thank you for your time.
 A: The $x$ coordinate of the vertex can be found by:
$$x = \frac{-b}{2a}$$
To find the $y$ coordinate, plug the $x$ you found above back into the original equation.So,it is:
$$f \left ( \frac{-b}{2a}\right)=a \frac{b^2}{4a^2} + b \left ( \frac{-b}{2a} \right) +c=\frac{b^2}{4a}-\frac{b^2}{2a}+c=c-\frac{b^2}{4a}$$
A: The formula $x=-\dfrac{b}{2a}$ can be derived using completing the square or differentiation (easied method).
The $y$-coordinate of the vertex can simply be found by plugging in the $x$-coordinate you found into your quadratic equation.

Completing the square:
$f(x)=ax^2+bx+c=a\Big(x^2+\dfrac{b}{a}x\Big)+c=a\Big(\Big(x+\dfrac{b}{2a}\Big)^2-\dfrac{b^2}{4a^2}\Big)+c=a\Big(x+\dfrac{b}{2a}\Big)^2-\dfrac{b^2}{4a}+c$
Hence, the vertex is $\Big(-\dfrac{b}{2a},-\dfrac{b^2}{4a}+c\Big)$
Differentiating:
$f(x)=ax^2+bx+c$
$f'(x)=2ax+b$
$f'(x)=0 \Longleftrightarrow x=-\dfrac{b}{2a}$
Find the $y$-coordinate by plugging $x=-\dfrac{b}{2a}$ into the original equation.
A: Memorizing formulas works really well in grade 5 and grade 6, but in grade 10, it just blows up in your face. Your teacher expects you to do some algebra to find the answer. The approach that I've taken uses linear transformations.
We know that
$$y=x^2$$
has a vertex at the origin [i.e. at $(0,0)$].
Stretching the curve by a factor of $a$ has no effect on the vertex (if you stretch zero, you get zero). The stretched equation becomes
$$y=ax^2$$
To shift every point up (including the vertex) by an amount of $k$, the equation would be
$$y=ax^2+k$$
Next we shift every point to the right by an amount of $h$
$$y=a(x-h)^2+k$$
So this new equation should have a vertex at $(h, k)$.
Expanding this equation we have
$$y=a(x^2-2hx+h^2)+k$$
$$y=ax^2-2ahx+h^2+k$$
$$y=ax^2+ (-2ah)x+(h^2+k)$$
we know that $b$ (aka $x$'s coefficient) is $-2ah$
$$-2ah=b$$
$$h=-\frac{b}{2a}$$
We also know that $c$ (aka the $x^0$'s coefficient) is $h^2+k$
$$h^2+k=c$$
$$k=c-h^2$$
$$k=c-\bigg(-\frac{b}{2a}\bigg)^2$$
$$k=c-\bigg(\frac{b}{2a}\bigg)^2$$
So the vertex is $\displaystyle\bigg(-\frac{b}{2a}, c-\frac{b^2}{4a^2}\bigg)$
