How do you find the vertex for #y=x^2+6x+7#?

Okay, so:
in order to get to the complete the square form, you need to halve the 'b' coefficient of your #ax^2±bx±c# equation. So, in this case, halve 6 to get:

#(x+3)^2#

right?

Let's expand the bracket, shall we?

#(x+3)^2#
#=x^2+6x+9#

Now, you'll notice that the 'c' isn't 7, but 9. That's where the 'n' comes in. The 'n' is supposed to be a simple calculation that gets your 'c' right; if your 'n' becomes a little bit too complicated, double check if you've done the halving/ expanding correctly!

#=x^2+6x+9#

We can get the 'c' to 7 by subtracting 2. The -2 becomes 'n'. So we get:

#=x^2+6x+9-2#
#=x^2+6x+7#

To get:

#(x+3)^2-2#

So now that you have your complete the square form done, think of what number gets ONLY the bracket to equal to 0. In this case, -3. Thus, -3 becomes your x value for the vertex.

#(-3+3)-2#

#=0-2#

However, you have the remaining 'n' part to think of as well, but since the bracket is already equal to 0,

#0-2=-2#

your 'n' value becomes the y value for the vertex! So the vertex of the graph is:

(-3, -2)

Hope this helps!

#y=x^2+6x+7# is a quadratic equation.

Graphing #x# and #y# on a cartesian plane would give a curve.
In this case, the graph is curving up due to the fact that #x^2# has a positive coefficient.

Since it is a graph curving up, we know that the gradient at points on the graph would be increasing from a negative number when the value of x increases.

At the vertex, or in other words, the minimum point (in this case since it curves up) has a gradient of 0.

Why?
Because the minimum point is a point where the gradient changes from negative to positive.

Firstly, to find its vertex, we need to form a gradient function, #dy/dx#

Differentiate #y=x^2+6x+7#
#dy/dx=2x+6#

We know that at vertex, gradient or #dy/dx=0#
#:.2x+6=0#

#x=-6/2=-3#

We have the x-coordinate of the vertex. To find the y-coordinate, simply plug in #x=-3# into #y=x^2+6x+7#

#y=(-3)^2+6(-3)+7#
#y=9-18+7=-2#

The vertex has a coordinate of #(-3,-2)#