What is the vertex of the parabola $v^2= 4c^2(u+c^2)$?

Question

How to calculate the vertex of the parabola when it's of the form $(x-h)^2=4c(y-k)$ ?

Also how to calculate the vertex of $u^2=1-2v$ ? (When there's a constant along with $v$.)

Answer 1

hint: write both equations in the standard form: $f(x) = y = ax^2+bx+c$, and the vertex $V = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right)$. The problem boils down...to find the $a,b,c$ for each case of the two equations.

Answer 2

Get your equation to form

$$(y-g)=t(x-h)^2$$ $$g,t,h\in const$$

Use the fact that

If we replace $x\to x-h$ graph shifts right by h.

If we replace $y\to y-g$ graph shifts g units down.

Try out, or click below

graph was $x^2=4cy$ (vertex of this graph is (0,0))$$$$which is shifted to right by $h$ and down by $k^2$ $$$$So vertex becomes $(h,-k^2)$

Answer 3

The vertex of a quadratic is the "tip" of the curve, e.g. the maximum or minimum point of which there is only one.

Hint: A quadratic function can be writen $y=ax^2+bx+c$. Completing the square gives

$$y=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a^2}.$$

So basically $$y=A(x+B)^2+C.$$

Hopefully you can see that, since $|A|(x+B)^2\geq 0$ for all $x\in\mathbb{R}$ then the vertex occurs at $(-B,C)$.

So if I were you I would expand your equations as efficiently as possible, complete the square, and write down the vertex.

Answer 4

For the first, it is nothing but $4 c y=x^2$ whose vertex is shifted to $(h,k )$.

Can you put the second into $y^2 = a( x-b)$ form ? Where you recognize $x$ shift as $b?$