problem give 2 points and we know that none of them are vertex,
how to find the equation
4 Answers
$\begingroup$Given two points on a parabola, $(x_1,y_1)$ and $(x_2,y_2)$ (where $x_1\neq x_2$), we try to find as much information as we can about the set of parabolas which pass through these points: $y=ax^2+bx+c$
Using the information given by those two points, this tells us the system of equations:
$$\begin{cases} ax_1^2+bx_1+c=y_1\\ ax_2^2+bx_2+c=y_2\end{cases}$$
Rewriting this in terms of a matrix equation:
$$\begin{bmatrix}x_1^2&x_1&1\\x_2^2&x_2&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}y_1\\y_2\end{bmatrix}$$
Solve the system using familiar methods (such as via row reduction)
For example, finding the set of parabolas which pass through the points $(0,1)$ and $(3,0)$ will produce the matrix equation:
$$\begin{bmatrix} 0&0&1\\9&3&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$
Row reducing the augmented matrix $\left[\begin{array}{ccc|c}0&0&1&1\\9&3&1&0\end{array}\right]$ yields $\left[\begin{array}{ccc|c}1&\frac{1}{3}&0&-\frac{1}{9}\\0&0&1&1\end{array}\right]$ implying that $c=1$ and $a=-\frac{1}{9}-\frac{1}{3}b$
Thus, all polynomials passing through those two points can be written as:
$$f(x)=(-\frac{1}{9}-\frac{1}{3}t)x^2+tx+1$$
for some $t\in\Bbb R$
$\endgroup$ 1 $\begingroup$If you are only required to find ONE possible parabola that suits the graph given, here is an approach.
Note that $x=4$ and for some $\alpha>-2$ are the roots of the parabola. Since the parabola opens downwards, we write $$y=-(x-4)(x-\alpha)$$
From the point $(-2,-2)$, $$-2=-(-6)(-2-\alpha)\implies\alpha=-\frac53$$
So the parabola can be
$$y=-(x-4)(x+\frac53)$$
To further verify the equation, we have $y=\frac{20}3>0$ when $x=0$, which is indeed true.
$\endgroup$ $\begingroup$If you have a parabola then it's a quadratic equation. You have a system of equations. Using f'' 's notation, $$ax_1^2+bx_1+c=y_1 \, \land \, ax_2^2+bx_2+c=y_2$$ 2 equations and 3 unknowns $a,b,c$.
$\endgroup$ $\begingroup$If the axis is vertical, we have three constants so we need to determine three constants in:
$$ y= a x^2 + b x + c $$
We require three points. Two points would not suffice.
For completeness..
If the axis is horizontal, we also find three constants in $x= a y^2 + b y + c. $
If the parabola axis is inclined to x-, y-, axes, then we need to find four constants in $y = a x +b + \sqrt{ c x +d }. $
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