This week in pre-calculus 11 I learnt how to find out an equation of a quadratic function from the characteristics that have been given. For example, when I have the vertex and only a point given and I have to find out an equation that satisfies both of those two characteristics. Or, when I only have the x-intercepts and the vertex provided. In both of these I am trying to find the stretch value, which is the number that is missing to complete the equation.

Here is an example:

The graph of a quadratic function passes through point B (3, 16), and the zeros of the function are 1 and 7. Write the equation of the function in general form.

Since the zeros of the function are just the x-intercepts of the graph, you have to use the factored form of the equation to figure out the missing value which is the stretch in this case. The factored form of the equation is: y =a(x – x1)(x – x2).
The first thing you have to do is substitute x1 = 1 and x2 = 7.

So, the equation so far looks like this:

Then you have to substitute the x = 3 and the y = 16 because it satisfies with the equation. Also, so that you can do algebra to figure out the missing value which is the stretch value “a.”

So, now the equation looks like this:

Now you have to solve for “a” which is the stretch value by using algebra:

Afterwards, when the stretch value is found this is how the equation looks like in factored form.

To get it to general form you have to use foil and distribute the -2 to the numbers inside the bracket:

Here is the final equation:

You can check it on Desmos and see that the parabola passes through the point (3, 16):

Here is another example involving finding the stretch value while only having the vertex provided and a point:

The coordinates of the vertex are V(4, 12) and the graph passes through A(7, 6):

Since you know the vertex, you have to use the standard or vertex form which is y = a(x – h)² + k to find the stretch value. The first thing you have to do is substitute the h = 4 and the k =12.

So, the equation will look like this so far:

Then you have to substitute the x = 7 and y = 6 because they satisfy the equation:

So now the equation will look like this:

Now you have to use algebra to find the “a,” which is the stretch value and the number that is missing in the equation:

This is how the equation looks like once you input the stretch value (in vertex/standard form):

You can double check on Desmos that the parabola passes through the point of (4, 12) and it does: