Next we will rewrite the left-hand side so that it can written as a square. To do that we have to add the square of half of the coefficient to x, which is ( 62)^2=9, to both sides of the equation. This is called completing the square.
The right-hand side of the function can be written as (x+3)^2+2. Finally, we have to make the right-hand side of our function match the right-hand side of the graphing form.
Graphing Form:& y=a(x- h)^2+ k
Our function:& f(x)=(x-( - 3))^2+ 2
b
For a function written in graphing form, the vertex is given by the point ( h, k).
f(x)=a(x- h)^2+ k
By examining our function we can identify this vertex. Notice that since a=1, a positive coefficient, the parabola will open upwards.
To complete the sketch we need to find more points.
c|c|c
x & x^2+6x+11 & f(x) [0.1em]
- 5 & ( - 5)^2+6( - 5)+11 & 6 [0.3em]
- 4 & ( - 4)^2+6( - 4)+11 & 3 [0.3em]
- 2 & ( - 2)^2+6( - 2)+11 & 3 [0.3em]
- 1 & ( - 1)^2+6( - 1)+11 & 6
Now we can complete the sketch.
c
The equation x^2+6x+11=0 asks when f(x) is equal to 0. If we examine the diagram from part B, we notice that y=0 and f(x) do not intersect.
Therefore, there is no value of x that solves the equation.
