# Vertex form of the function

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Commented:
A very different approach (that may be much easier to program is:

Take the first derivative of the function
Solve it for 0 (i.e. what x gives you a first derivative =0?)
Plug that x in the original function to get the value of y
Commented:
Depending on the form of the original function, this may be very straightforward.
Commented:
Were you needing alternatives for the other solutions?  The intercepts should be easy.

For the min/max you can do the second derivative of the original function at the vertex.  If it is negative then the vertex is a maximum, a positive indicates a minimum, and 0 indicates (I forget the proper name) that it is a "flat spot with one side increasing and the other decreasing".

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Author Commented:
Can you please show me a little bit work as I am not in touch with calculus since a long time.
I think first derivative of that equation would be:
2x-8+0
2x=8
x=4
The original function is:
x^2-8x+7=0
substituting x
8-32+7=0
y=31
I am not sure if this is correct. or I may mixing the integration with derivation.
Commented:
x^2-8x+7=y is:
16-32+7=y
-9=y
Author Commented:
IS THIS IS CORRECT

Second derivative of the same equation:
x^2-8x+7=0

I think x=0

substituting x in that equation
y=7
Commented:
The second derivative of x^2-8x+7 is the first derivative of the first derivative.

So... the first derivative is 2x-8

The second derivative is 2.

Since the second derivative is positive, then the point is a minimum.  That is, the function slopes upward on either side of that point.

Note that the second derivative doesn't depend on x.  That means that as you move from left to right on the graph, its slope increases.  When the slope is negative (as in to the left of the vertex), "slope increases" means it becomes less negative.
Author Commented:
Thank You.
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