Question

Find

Answer 1

wolfram tells that its range lies between[2,0]
but how

Answer 2

*-2

Answer 3

Factor it first...

Answer 4

\[\LARGE - \sqrt{-(5+6x+x^2)}\]

Answer 5

for range put y = f(x)

Answer 6

Oh, range not domain

Answer 7

ok... so look at the function...


\[y = - \sqrt{-(5 + 6x + x^2)} = -\sqrt{-(x + 5)(x + 1)} \]

so the vertical asymptotes exist at x = -5 and -1 which are the zeros.


find the line of symmetry

x = -6/(2*1)

so x = -3

substitute it and you'll find y = -2

so I'd say the range is [-2, 0]

Answer 8

but then who an I to contradict Wolfram Alpha...

Answer 9

@campbell_st he corrected himself above, it was [-2,0]

Answer 10

only one thing line of symmmetry =?

Answer 11

yep... for parabola in the form

\[y = ax^2 + bx + c\]

the line of symmetry is

\[x =\frac{-b}{2a}\]

Answer 12

kk
now i got it @campbell_st
thank you so much

Answer 13

If you're asking what it is... http://2012books.lardbucket.org/books/elementary-algebra/section_12/21bb411a3f34e1fbb8c80e088876fcbf.jpg

And it's where the max or min value of the parabola occurs.

Answer 14

And it's halfway between the x intercepts, so if you know those, you can easily find the axis of symmetry.

Answer 15

you might notice its part of the general quadratic formula...

but as @agent0smith said , for a parabola... the max or min value lies on the line of symmetry

Answer 16

and you understand why the asymptotes occur....

Answer 17

yes

Answer 18

one thing how u make a graph, it is very clear

Answer 19

https://www.google.com/search?q=-sqrt(-(5%2B6x%2Bx%5E2)+)&aq=f&oq=-sqrt(-(5%2B6x%2Bx%5E2)+)&aqs=chrome.0.57j62l3.275&sourceid=chrome&ie=UTF-8

this is the graph of your function, so you can actually see the range (and domain)

Answer 20

its easy... its a parabola with zeros at -5 and -1 which are also asymptotes...

its concave up with a minimum value of -2