0
$\begingroup$

Given a parabola with this equation:

$$y = x^2 + bx + c$$

And given 2 consecutive points of the parabola:

$$p_1 = (x_1, y_1), ~ p_2 = (x_2, y_2), ~x_2=x_1 + 1$$

Is there a way to find the vertex of the given parabola?:

$$v = (x_v, y_v) $$

PD: In the parabola equation, the only unknown is "b", and all variables (except the vertex) are given integers.

Update:

If the parabola equation is:

$$y = ax^2 + bx + c$$

and "a" is known. What changes in the solution?

$\endgroup$

2 Answers 2

0
$\begingroup$

$\displaystyle y = x^2 + bx + c$

or, $\displaystyle y = \left( x + \frac{b}{2} \right)^2 + \frac{4c - b^2}{4}$

Hence

$\displaystyle x_v = -\frac{b}{2}$

and

$\displaystyle y_v =\frac{4c - b^2}{4}$

Our only task is to compute $b$ in terms of $x_1, x_2, y_1, y_2$ and $c$

Since

$\displaystyle y_1 = x_1^2 + bx_1 + c$

and

$\displaystyle y_2 = x_2^2 + bx_2 + c = (x_1+1)^2 + b(x_1+1) + c$

Subtracting,

$y_2 - y_1 = 2x_1 + 1 + b$

or, $b = y_2 - y_1 - 2x_1 - 1$

and we are done!

Please let me know if that is what you wanted.

$\endgroup$
5
  • $\begingroup$ Good answer! But, I had the need to update the question. I have found another similar problem to solve. With this addition the question will be answered. Sorry for the inconvenience $\endgroup$ Commented Aug 19, 2019 at 0:01
  • $\begingroup$ Please post a new question. $\endgroup$
    – PTDS
    Commented Aug 19, 2019 at 0:05
  • $\begingroup$ Ok. But how you obtain the second formula: y = (x + b/2)^2+ (4c - b^2)/4 ? $\endgroup$ Commented Aug 19, 2019 at 0:08
  • $\begingroup$ Observe the expression $x^2 + bx + \ldots = x^2 + 2*x*\frac{b}{2} + \ldots$ If you try to "complete the square", what should you do? Just add and subtract $\left( \frac{b}{2} \right)^2$. Is it clear now? $\endgroup$
    – PTDS
    Commented Aug 19, 2019 at 0:14
  • $\begingroup$ It's very clear. Thanks! $\endgroup$ Commented Aug 19, 2019 at 0:16
0
$\begingroup$

Since you only have one unknown, $b$, then you can use one of the two points you know the parabola passes through (say $p_1$) to solve the equation $y_1 = x_1^2 + bx_1 + c$ where the unknown is just $b$.

Once you have that you know that the coordinates of the vertex are $(x_v,y_v)$ for: $$ y= (x−x_v)^2+y_v\,, $$ obtained from your $y= x^2 + bx + c$ by re-arranging terms.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .