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3 votes
X = x^2 + 4x + 4

Axis of Symmetry:
Vertex:
Domain:
Range:
And I also need the graph points


Please show your work and how you got it.

1 Answer

3 votes
You mean Y=x^2 + 4x + 4, this is quadratic function (convex parabola) which standard form is y=ax^2+bx+c

1. First we will find zero of the function, y=0 or as you like intersection
with x axis.
y=0 =>. x^2+4x+4=0 => this is the square of the binomial ( x + 2)^2 = 0
=> x + 2 = 0 => x= -2 In this case it's a touch point with x axis.
2. Intersection with y axis => If x=0 => y=4 (0, 4)
3. The coordinate of vertex are x= - b/2a = - 4/2 = -2 and y=(4ac-b^2)/4a
y= (4* 1*4 - 4^2)/ 4 = (16-16)/4= 0 Vmin ( -2, 0) this point is at same time zero of the function.
4. Domain:. Dx=R (field of real numbers)
5. Range: y belong to interval ( - infinite, 0) or R-
6. Function's sign y>0 => x belong to intervals (-infinite, -2) U ( -2, +infinite)
7. Function's growth or decline
for x belong interval ( -infinite, -2) => function decline
for x belong interval ( -2, +infinite) => function is growing

Thats all !!!!!!






User Horatio
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