how to create a visual model (x+3)^2

Answer 1

y = (x+3)^2 is an upward opening parabola with vertex = minimum point = (-3,0) = x intercept where the parabola is tangent to the x axis

axis of symmetry is x=-3

y intercept = 9 or the point (0,9)

y= x^2 +6x +9 = 9 when x=0

the parabola is downward sloping for x<-3 and upward sloping for x>-3

it has zero slope at x=-3

the parabola is concave upward everywhere, for all values of x. Graphically it's U shaped

there is only one local minimum = global minimum = 0'

there is no local maximum or global maximum, as x approaches negative or positive infinity, y approaches positive infinity

its parent function is y=x^2, it's shifted horizontally to the left by 3 units. every point on the parabola y=x^2 is shifted left by 3 units

its focus is directly above the vertex with coordinates (-3, 1/4)

directrix is directly below the vertex: y= -1/4

the vertex is midway between the focus and the directrix

every point (x,y) = (x, x^2+6x+9) is equidistant from the focus and the directrix

for example (-5/2, 1/4) is 1/2 unit from the focus (-3, 1/4) and 1/2 unit from the line y=-1/4