Question

y=x^2-7x+12
min or max? and value? how is this determined ??

Answer 1

what is the "leading coefficient' i.e. the coefficient of the term \(x^2\)?

Answer 2

so if it is positive it is considered min? @satellite73

Answer 3

the leading coefficient is positive, so the parabola opens up, and so it has a minimum value (the second coordinate of the vertex) but no maximum value

Answer 4

thanks I get it, so how am i supposed to find the value ? @satellite73

Answer 5

first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \(\frac{7}{2}\)
second coordinate of the vertex is what you get for \(y\) when you replace \(x\) by \(\frac{7}{2}\)

Answer 6

thank you, you have been such a help! @satellite73

Answer 7

You could also find the derivative of it in this case it is \[2x - 7\] and then solve for when it equals 0 so you get \[0 = 2x - 7\] so x = 7/2 plug 7/2 back into the orginal equation so you can find the y- coordinate so \[(\frac{ 7 }{ 2 })^2 - 7(\frac{ 7 }{ 2 })+12=y\]
if you simplify this you get y = \[-\frac{ 1 }{ 4 }\]
so the coordinate of the vertex is \[(\frac{ 7 }{ 2 },-\frac{ 1 }{ 4 })\]
So that gives you the coordinate of the vertex. Now to find out if it is a minimum or maximum take the derivative of the derivative so the derivative of \[2x- 7\]
which is 2. If the derivative is positive which it is the vertex is a minimum if negative it is a maximum. So the point \[(\frac{ 7 }{ 2 }, -\frac{1 }{ 4 })\] is a minimum

This method uses calculus I'm not sure if u know it yet.
The derivative is the slope at any point so in the method i calculated when the slope (derivative) was 0 which is either a maximum or a minimum, then i calculated the x and y coordinates by solving for x and substituting for y, and finally i took the derivative of the derivative (slope of the slope) to find if it was negative or positive. The negative means maximum positive means minimum. In this case it was positive so it was a minimum.

p.s. i think this Calculus method only works with quadratic equations emphasis on i think :)