Maths graphs

why does putting an equation into completed square form show the vertex of the graph?

So for example:
f(x) = x² + 4x + 3
In completed square form is x = -2 +- root1 (I think)
And from this the vertex is (-2,-1)

But why?

why does putting an equation into completed square form show the vertex of the graph?

So for example:
f(x) = x² + 4x + 3
In completed square form is x = -2 +- root1 (I think)
And from this the vertex is (-2,-1)

But why?

I am afraid you didn't complete the square properly.
the ans is ((x+2)^2) -1

How did you get that? this is what I did:

x^2 + 4x + 3 = 0
(x+2)^2 -4 + 3 = 0
(x+2)^2 - 1 = 0
(x+2)^2 = 1
x+2 = +-root 1
x= -2 +- root 1

which bit was wrong?

...

I am afraid you didn't complete the square properly.the ans is ((x+2)^2) -1

As Azhar said, the square wasn't completed properly. But imagine you had the graph y = f(x) where f(x) = x²

Have you learnt graph translations?

What if we had y = f(x+2)? This means f(x+2) = (x+2)² !

What does f(x+2) mean from the f(x) graph? It means that the graph is shifted 2 units to the left.

So what if we plonk on your -1?

y = f(x+2) - 1

this means we get y = (x+2)² - 1, which happens to be your graph.

What does a translation of y = f(x+2) - 1 represent? It represents shifting to the left 2 units, as we saw earlier. But NOW we have this -1 outside. So what happens if we take away 1 from all f(x+2) values? The graph shifts down by 1 unit.

So in your example, the y = x² graph shifted 2 units left and 1 unit down where the original vertex was the origin. What happens to the point (0, 0) when shifted 2 left and 1 down? It becomes (-2, -1)

I'll complete the square correctly for you: $x^2 + 4x + 3 = (x+2)^2 - 1$.

Now, since you know that square quantities are always $\geq 0$ then the minimum value of this expression occurs when the squared term is zero.

This means that $\min(x^2 + 4x + 3) = \min((x+2)^2 - 1)$ occurs precisely when $(x+2)^2 = 0 \Rightarrow x = -2$.

The minimum value of function at the point $x=-2$ is $\min (x^2 + 4x + 3) = \min((x+2)^2 -1 ) = 0-1 = -1$.

Damn, you win this one. At least our explanations are substantially different to warrant the ninja.

7.5/11

And yeah! I think yours is a lot simpler though And ofc in maths simpler = better

Daaamn, getting there. K : D ratio

Nah, I think yours is just as nice (if not nicer), transformations can be a very elegant thing!

That's not the vertex, that's you finding the roots of the quadratic. Oh, and $\sqrt{1} = 1$.

Also: moved to maths.

Damn, you win this one. At least our explanations are substantially different to warrant the ninja.

Oops Thanks for your help

How did you get that? this is what I did:

x^2 + 4x + 3 = 0
(x+2)^2 -4 + 3 = 0
(x+2)^2 - 1 = 0
(x+2)^2 = 1
x+2 = +-root 1
x= -2 +- root 1

which bit was wrong?

lol, you guys