SOLUTION: I have one more problem that I need help in solving....I hope someone can help me. Thank you for your time and for all your help! y=x^2+4x+5 Please include: a) opens up, dow

Question 43904This question is from textbook College Algebra
: I have one more problem that I need help in solving....I hope someone can help me. Thank you for your time and for all your help!
y=x^2+4x+5
Please include:
a) opens up, down, right or left
b)vertex
c) axis of symmetry
d) X-intercept
e) y-intercept
f)graph
Thanks a bunch again! smccollough1
This question is from textbook College Algebra

Answer by stanbon(75887) (Show Source): 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:

Cheers,
stan H.

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