Is there a way to find the vertex of a parabola, given 2 consecutive points of the parabola?

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?

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

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.

• 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 Commented Aug 19, 2019 at 0:01
• Please post a new question.
– PTDS
Commented Aug 19, 2019 at 0:05
• Ok. But how you obtain the second formula: y = (x + b/2)^2+ (4c - b^2)/4 ? Commented Aug 19, 2019 at 0:08
• 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?
– PTDS
Commented Aug 19, 2019 at 0:14
• It's very clear. Thanks! Commented Aug 19, 2019 at 0:16

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.