I think it could be a) no real solutions?
Am I correct? If not, please explain...
How many real solutions does the function shown on the graph have?
a) no real solutions
b) one real solution
c) two real solutions
d) cannot be determined
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y = (x+2)^2 + 3 is shown on the graph.
I am having trouble figuring this one out. I have been looking in my book for a while now. Somebody please help, it would be very much appreciated!
5 answers
Whoops. I meant the graph was showing: y=(x+2)^2+2
I was thinking that it is a) because it does not have any x-intercepts?
( x + 2 )² + 2 = 0
Subtract 2 to both sides
( x + 2 )² + 2 - 2 = 0 - 2
( x + 2 )²= - 2
Take square root of both sides
x + 2 = ± √( - 2 )
x + 2 = ± √( - 1 ∙ 2 )
x + 2 = ±√( - 1 ) ∙ √2
x + 2 = ± i √2
Subtract 2 to both sides
x + 2 - 2 = ± i √2 - 2
x = - 2 ± i √2
The solutions are:
x = - 2 + i √2 and x = - 2 - i √2
The function ( x + 2 )² + 2 no real solutions.
Answer a) is correct
Subtract 2 to both sides
( x + 2 )² + 2 - 2 = 0 - 2
( x + 2 )²= - 2
Take square root of both sides
x + 2 = ± √( - 2 )
x + 2 = ± √( - 1 ∙ 2 )
x + 2 = ±√( - 1 ) ∙ √2
x + 2 = ± i √2
Subtract 2 to both sides
x + 2 - 2 = ± i √2 - 2
x = - 2 ± i √2
The solutions are:
x = - 2 + i √2 and x = - 2 - i √2
The function ( x + 2 )² + 2 no real solutions.
Answer a) is correct
Just looking at the equation, we can tell that the vertex is (-2,2) and the parabola opens upwards. So clearly, it cannot cross the x-axis. So clearly, no real solution
3 answers