You can
put this solution on YOUR website!y=x^2+4x+5
Write in vertex form by completing the square, as follows:
x^2+4x=y-5
x^2+4x+4=y-5+4
(x+2)^2=y-1
Please include:
a) opens up, down, right or left
Has the form y=+x^2 s]o it opens up.
b)vertex
vertex = (-2,1)
c) axis of symmetry
vertical line through the vertex: x=-2
d) X-intercept
Let y=0 and solve for x in original equation:
x^2+4x+5=0
x=[-4+sqrt(16-20)]/2; discriminant is negative so no x intercepts.
e) y-intercept
Let x=0 in original equation: y=5
f)graph
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
The discriminant -4 is less than zero. That means that there are no solutions among real numbers.
If you are a student of advanced school algebra and are aware about imaginary numbers, read on.
In the field of imaginary numbers, the square root of -4 is + or -  .
The solution is 
Here's your graph:
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Cheers,
stan H.
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