Variation in graph with function f(x) = x^2 + kx +lposition (o,o) with parabola graph.Discribe the position of this graph f(x+c)

The given equation of parabola is given in standard
form.

Since the leading coefficient is >0, the
parabola opens upwards.

The axis of symmetry of given
parabola is parallel to y axis and it is given by:

x =
-k/2*1

x = -k/2

We'll write
the quadratic function f(x+c) = (x+c)^2 + k*(x+c) +
l

f(x+c) = (x+c)^2 + k*(x+c) +
l

f(x+c) = x^2 + 2cx + c^2 + kx + kc +
l

f(x+c) = x^2 + x(2c + k) + c^2 + kc +
l

Again, the leading coefficient is >0, therefore
the parabola opens upwards.

The axis of symmetry of given
parabola is parallel to y axis and it is given by:

x = -(2c
+ k)/2

x = -2c/2 - k/2

x =
-k/2 - c

We'll consider the
example:

f(x) = x^2 + 6x + 7

x
= -6/2

x = -3

f(x+2) = x^2 +
x(2*2 + 6) + 2^2 + 6*2 + 7

f(x+2) = x^2 + 10x +
23

x = -10/2

x =
-5

We can see that the vertex of parabola was moved to the
right by the amount of c = 2.

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src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,x^2 +
10x+23,null,0,0,,,orange,1,none,func,x^2
+6x+7,null,0,0,,,green,1,none"/>

We notice that the
green parabola is represented by the function f(x) = x^2 + 6x + 7, having the x
coordinate of vertex, x = -3 and the orangle parabola is represented by the function
f(x) = x^2 + 10x + 23, having the x coordinate of vertex, x =
-5.

The axis of symmetry is displaced by the
amount -c, therefore the vertex of parabola is moved to the left, with the amount
-c.