How to find the minimum point on a quadratic curve?

Or highest point for that matter?

E.X: I have the equation y=x^2-8x+23

2. Originally Posted by Mukilab
How to find the minimum point on a quadratic curve?

Or highest point for that matter?

E.X: I have the equation y=x^2-8x+23

Hi Mukilab,

This curve will have a minimum because the coefficient of the quadratic term is positive.

The minimum point is the vertex.

You can find the x-coordinate of the vertex by using this little formula:

$x=\frac{-b}{2a}$

Once you have the x-coordinate of the vertex, plug it into the quadratic function to find the corresponding y-coordinate.

You could also complete the square to find the vertex, but who wants to do that, right?

3. Originally Posted by Mukilab
How to find the minimum point on a quadratic curve?

Or highest point for that matter?

E.X: I have the equation y=x^2-8x+23
Take the derivative and equate to 0.

$y^{ \prime } = 2x - 8 = 0$

Min happens at

$x = 4$

If we don't use derivatives (which I suspect you are not supposed to) we complete the square.

$y = x^2-8x+23 = (x^2 - 8x + 16 - 16) + 23 = (x-4)^2 + 7$

From here we can find the max/min. Since this parabola is facing upwards, we know there in a minimum.

This is known as "vertex form" and the min/max points happen at the vertex. In this case,

the vertex is $(4,7)$ which comes from the negative of the number inside the bracket with the x, and the number that is added.

In general

$y = (x-p)^2 + k$

has the vertex

$(p,k)$

4. Originally Posted by AllanCuz
Take the derivative and equate to 0.

$y^{ \prime } = 2x - 8 = 0$

Min happens at

$x = 4$

If we don't use derivatives (which I suspect you are not supposed to) we complete the square.

$y = x^2-8x+23 = (x^2 - 8x + 16 - 16) + 23 = (x-4)^2 + 7$

From here we can find the max/min. Since this parabola is facing upwards, we know there in a minimum.

This is known as "vertex form" and the min/max points happen at the vertex. In this case,

the vertex is $(4,7)$ which comes from the negative of the number inside the bracket with the x, and the number that is added.

In general

$y = (x-p)^2 + k$

has the vertex

$(p,k)$
Surely its far easier to use the method master stated? Or does the one you have stated work well under different circumstances?

5. Originally Posted by Mukilab
Surely its far easier to use the method master stated? Or does the one you have stated work well under different circumstances?
Completing the square takes 15 seconds and is a requisite skill for higher level math (i.e. derivatives and integrals). I would learn to master it as it will help you on your current exams (for other problems not of this nature and when Master's equation does not apply) and in the future.

6. Do not know what the derivative is. Do no know what 'equate' means, is this change it to zero and change others as well in ratio?

what is y^{/prime}??????

Lastly, why did you add in +16, -16? Somehow you changed -8x, +16, -16 all into -4 and changed 23 into 7?!

7. Originally Posted by Mukilab
Do not know what the derivative is. Do no know what 'equate' means, is this change it to zero and change others as well in ratio?

what is y^{/prime}??????

Lastly, why did you add in +16, -16? Somehow you changed -8x, +16, -16 all into -4 and changed 23 into 7?!
This is all done in the process of 'completing the square'.
If you haven't studied how to do this, then go back to
the way I showed you.

8. Originally Posted by masters
This is all done in the process of 'completing the square'.
If you haven't studied how to do this, then go back to
the way I showed you.
I need this for another question where someone told me to 'complete the square' and redirected me here

9. Originally Posted by Mukilab
I need this for another question where someone told me to 'complete the square' and redirected me here
Maybe this will help: How to Complete the Square