Find the coordinates of the vertex of the quadratic function y= x^2 + 4x + 6?

You can do it by completing the square.

y = x^2 + 4x + 6

Separate the first two terms.

= (x^2 + 4x) + (6)

Take half of the 4 (= 2) and square it (2^2 = 4).

Add this 4 to the first bracketed term and subtract it

from the second bracketed term.

= (x^2 + 4x + 4) + (6 - 4)

= (x + 2)^2 + 2

= [x - (-2)]^2 + 2

This is now in the form of the vertex formula,

(x - h)^2 + k, where (h,k) is the vertex.

Here, h = -2 and k = 2, so the vertex is at the point (-2, 2).

1. Since the function is a parabola going up and down (since x2 is negative we know it opens down), the vertex is going to be the maximum value since everything else is below it. So we find the vertex: Plug in points of x - you're looking for the highest possible value. Try x=-1 and you get y=-1. Then try x=0 and you get y=-1. You got the same answer so the top is probably right in the middle of these x-values. Try x=-1/2 and you get y=-3/4. This is your vertex and max point (-0.5, -0.75) 2. Try x=0 and y=9; x=1 and y=6; x=2 and y=9 Right in the middle of x=0 and x=2 (they have the same y-coordinate so the vertex is in the middle) is x=1. So your vertex, which is the minimum point because the 3x^2 is positive (so everything is above it) is (1,6) 3. All you have to do is graph it. If the graph hits the x-axis once, it has one answer. If it hits twice, two answers. If it doesn't hit at all, it has no real answers. When you graph it you see that it hits only at x=-2. Therefore it has one real solution at x=-2. 4. y=x^2+4x+8 complete the square: (x - h)^2 means everything with an x has to be inside ( ). y= (x^2+4x+4) - 4 + 8 y= (x+2)^2 + 4 y= 1(x+2)^2 + 4 is your answer

y=x^2+4x+6

y-6=x^2+4x by completing the square

y-6+4=x^2+4x+4

y-2=(x+2)(x+2) simplify

y-2=(x+2)^2

y=(x+2)^2+2

The vertex form of a quadratic is given by y = a(x – h)2 + k, where (h, k) is the vertex.

therefore: the vertex is (-2,2)

The vertex will be on a point such that

dx/dy=0

if F(x)=x^2+4x+6

dx/dy=2x+4=0 -for the vertex

x=-2 , y=2

So the vertex is in (-2,2)

y=x^2+4x+6

=x^2+4x+4+6-4=(x+2)^2+2

the vertex is(-2,2).answer

find dy/dx =0

2x+4 =0

x= -2

so y=(-2)^2+4(-2)+6 =2

vertex is(-2,2)