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I'm not the best at algebra so I will try to word this in the best way I can.

Let's say I have 2 solutions in each of 2 solution sets:

Set 1: (6, 0) and (18, 0) (visual here)

Set 2: (0, 0) and (2, 0) (visual here)

As you can see, the y position of the vertices for both of the equations shown above are different (set 1's vertex has a y position of about -36 and set 2's vertex has a y position of about -1)

How would I go about writing a quadratic equation that incorporates 2 solutions (for example 6 and 18 as shown in set 1) and makes the y position of the vertex always be at y = 10? Again I'm looking for an equation that will do this for any 2 solutions.

If you would like more clarification, ask and I will try to provide some more information. Any help is appreciated!

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    You just use the fact that you can write the quadratic as $a(x-x_1)(x-x_2)$ and then find $a$ using the conditions you set – Triatticus Oct 14 '17 at 02:19

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Let's define some terms that will help with the question.

The roots or zeros of a polynomial of degree $n$ are the solutions to the polynomial. On a graph, they are shown as the $x$-intercepts.

The 'vertex' of a quadratic graph is called a turning point, which can be minimum or maximum. It is the point which the gradient is equal to zero.

A useful form of writing a quadratic is called the turning point form, which is what you need at the moment. Turning point form is said to be written in the form:

$$y = a(x-h)^2+k$$

where $a$ measures the width of the parabola, and $(h,k)$ is the turning point of the graph.

If you are looking for a quadratic which has its turning point at $y=10$, then any quadratic will be in the form

$$y = a(x-h)^2 + 10$$

Any real value of $h$ will work, but $a$ must be negative - this reflects the graph along the x-axis. By doing this, the graph is upside-down (we call this concave down) and there will always be two solutions.

Therefore, to complete your problem, a quadratic in the form $y=a(x-h)^2+k$ such that $a<0$, $k=10$ will give a turning point at $y=10$ with two distinct solutions.

You can always expand the brackets to get to the normal quadratic form $y = ax^2 + bx + c$.

Let's do an example: let $a=-2$ and $h=3$ and $k=10$.

then $$y=-2(x-3)^2+10$$

the graph of $-2x^2+12x-8$ looks like this.

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