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I was originally given the value $(4,-2)$ as the vertex of a parabola and told that it also includes the value $(3,-5)$. From this point, I deduced that the next point would have the same y-value as the point whose x-value is equidistant from the vertex, so the next point would be $(5,-5)$. I also know that since the parabola's vertex is higher than the two values surrounding it, it is a negative parabola. I am now stuck and do not know how to continue about finding the equation for this limited input/output table.

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Any parabola can be written as:

$ y = Ax^2 + Bx + C$

As you said, our parabola passes through:

$(4,-2)$, $(3,-5)$ and $(5,-5)$

Substituting these values into the general equation of the parabola we get:

$-2 = 16A +4B +C$

$-5 = 9A + 3B +C$

$-5 = 25A + 5B + C$

So we have to solve this 3 by 3 linear system. The solution is:

$A = -3 \quad \quad B = 24 \quad \quad C = -50$

The final answer is then:

$y = -3x^2 + 24x - 50$

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Vertex form: $g(x) = a(x-h)^2+k$ where $(h,k)$ is your vertex.

Here $$g(x) = a(x-4)^2-2 $$ and $$g(3) = -5 \Rightarrow -5 = a(3-4)^2-2 \Rightarrow -3 = a \Rightarrow g(x) = -3(x-4)^2-2.$$

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I see @Rod gave an answer using the fact that the vertex was given. I'd like to give another method that can be used to find the parabola given any three arbitrary points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$.

Assume $y = ax^2 + bx +c$ and generate a system of three equations. I'll put it in matrix form for brevity $$ \left(\begin{array}{ccc} x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\\ x_3^2 & x_3 & 1\\ \end{array}\right) % \left(\begin{array}{c} a\\ b\\ c\\ \end{array}\right) % = \left(\begin{array}{c} y_1\\ y_2\\ y_3\\ \end{array}\right) $$

To solve for the coefficients, just multiply by the inverse of the square matrix. In this problem we get

$$ \left(\begin{array}{c} a\\ b\\ c\\ \end{array}\right) % = % \left(\begin{array}{ccc} 16 & 4 & 1\\ 9 & 3 & 1\\ 25 & 5 & 1\\ \end{array}\right)^{-1} % \left(\begin{array}{c} -2\\ -5\\ -5\\ \end{array}\right) % = % \left(\begin{array}{c} -3\\ 24\\ -50\\ \end{array}\right) $$ So therefor $y= -3x^2 + 24x - 50$ represents the equation for our parabola.

This clearly isn't the most effective method in this case, but perhaps it would be helpful for similar problems with arbitrarily chosen points.

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you may also usefully treat this as an exercise in change of coordinates. suppose we shift the origin to the vertex, and reflect through this new origin. the transformation is $$ X = 4- x \\ Y = -2 -y $$ the point $(3,-5)$ becomes $(1,3)$ so the equation in the transformed co-ordinates is: $$ Y=3X^2 $$ i.e. (going back to the original co-ordinate system) $$ -2-y = 3(4-x)^2 $$ or $$ y = -3x^2 + 24x -50 $$

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Note that two points are not sufficient alone to determine a unique parabola (just as two points do not determine a unique circle). However, if you also assume that the parabola is "vertical", the extra condition does uniquely determine the parabola.

Assuming the parabola is "vertical", then it is the graph of an equation of the form $$y=c(x-a)^2+b$$ where the vertex is the point $(a,b)$. You are told that $a=4$ and $b=-2$, so the equation is of the form $$y=c(x-4)^2 -2$$

Plug in the other point known to lie on the parabola to find $c$: $$-5 =c(3-4)^2-2$$ $$-5=c-2$$ $$c=-3$$

So an equation of the parabola is $$y=-3(x-4)^2-2$$

This can be "simplified" if desired to get $$y=-3x^2 +24x-50$$

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