Find the Coordinates for the equations minimum point

y=x^2-4x +3

Question

Find the Coordinates for the equations minimum point

y=x^2-4x +3

anonymous · Answer

can you use calculus?

anonymous · Answer

This should be posted in the math section, but Complete the Square to get the minimum point

anonymous · Answer

lmfao opps
you are in the math section

anonymous · Answer

ugh i thought i was in physics

anonymous · Answer

ha! nope but can you help anyways?

anonymous · Answer

@ihatephysicslol

anonymous · Answer

$$y = x^2 - 4x + 3 $$  which essentially is, $$y = Ax^2 + Bx $$ okay so in order to find the minimum you would have to complete the square... so basically all you do is take your A and B value and make a perfect square. How you do that is by isolating your A and B first so.. 
$$y= (x2−4x) +3$$
then taking your B value... $$Cnew =( \frac{ B }{ 2 }) ^2 $$
this will give you your new C value for the isolated part of your equation.. so... 
$$y=(x2−4x + (Cnew))+3$$

With the new C value, if we add it into the equation like this $$y=(x2−4x + Cnew - Cnew)+3$$ we are basically adding zero... okay so then to make like easier we try to get rid of the negative Cnew by expanding the bracket and add 3 to it so it looks like this... $$y=(x2−4x+Cnew)+3+(Cnew) $$
it looks a little bit weird but once you have an actual number for Cnew it will look normal :p 
okay now with the equation that we've created inside the bracket, you would simply just factor it and it will become a perfect square and you'll have your final equation which should look a little something like this but with numbers $$y = A(x-h)^2 + k   $$ where (h,k) is basically (x,y) and your "k" or "y" in this case would be your minimum value. Also note that $$(x-h)^2$$because "h" is a negative inside the bracket it will become a positive when you put it into your (h,k) brackets, and vice versa

anonymous · Answer

oopps $$y=Ax2+Bx + C $$ *** forgot the C in the beginning !!

anonymous · Answer

$$y=Ax^2+Bx+C$$ and the two is a square ... these equation things are so frustrating lol

anonymous · Answer

$$y=(x^2−4x+Cnew)+3+(-Cnew)$$

ahh i forgot the negative sign on the Cnew when we expanded it... omg i'm sorry for all these little mistakes