how can I find the domain and the range of this function: y=x^2+8 and y=Squareroot 5-x?

Question

please explain step by step on how to find the range and the domain and how to determine of the equation defines y as a funsction of x.

Answer 1

y is a function of x iff for each member of the domain there exists exactly one solution. IE, if you can take f(8), for example and get two different answers, it is not a function. Graphically, if you draw a vertical line through a point on the x-axis, it will intersect the curve in 0 or 1 point.

domain of y=x²+8. What will cause an illegal operation (division by zero or √ of a negative number)? Nothing. Therefore the domain is {x|x is any real number}

The range is what y can be. Range is {y|y ≥ 8} or y is on [8,∞) if your teacher prefers interval notation.

y=√(5-x) {I'm assuming that this is how it's supposed to look}
Domain. 5-x ≥ 0.
-x ≥ -5
Remember to reverse the inequality when multiplying by -1.
x≤ 5
Domain (-∞, 5]

Range is [0,∞)

Answer 2

1) y = x^2 + 8

The domain is easy; it is all real numbers. Generally, a polynomial's domain is all real numbers.

As for the range, it is worth nothing that this is a parabola. Normally, when we have a parabola in vertex form, it will look like this:

y = a(x - h) + k

where the coordinates of the vertex are (h, k). In our case, our function can be rewritten as

y = 1(x - 0) + 8

This shows the coordinates of our vertex to be (0, 8).
This tells us the range would be [8, infinity), or, in set notation, the set of all x such that x is greater than or equal to 8.

2) y = sqrt(5 - x)

One thing to identify in obtaining the domain is that the values of x are restricted to the square root function. That is, the inside of the square root MUST be greater than or equal to 0. Writing that in the form of an inequality,

5 - x >= 0

Now, just solve this inequality.
-x >= -5

And now, multiplying both sides by -1 will flip the inequality sign, and

x <= 5

Therefore, the domain of this function in set notation is:
{x | x <= 5} "The set of all x such that x is less than or equal to 5."

In interval notation, this would be (-infinity, 5].

To find the range, note that the result of the square root will always be greater than or equal to 0. Therefore, our range would be

{y | y >= 0}; in interval notation, [0, infinity)

Answer 3

y=x^2+8
x can be any real number so domain is all real numbers
y has a minimum value of 8, so y=>8 is the range

y=Squareroot 5-x
x must be > 5 or y becomes imaginary and must be excluded.
so domain is x =< 5.
y has a minimum value of 0 when x = 5.
As x --> - infinity, y becomes larger and larger without limit
Therefo range is y > 0.

Answer 4

y=x^2+8

here is the hint, every time you see a quadractic, the domain will be always real numbers.

Range: when you sub 0 for x, y is 8
when you sub -1, y= 9
whyn you sub 1, y=9
therefore, y>/8


y=sqr(5-x)
the domain is what makes the quanlity under the root greater than or equal to 0
5-x>/0
-x>/-5
x
therefore, the domain is x must be less than or equal to 5

Range: since the is a positive sign infront of the root
y must be greater than or equal to some thing

So you partly have: y>/ ?

what happened if you put 5 for x? you get 0 right?

therefore: y>/0

Answer 5

if I remember it correctly, domain represent the value set x can have, range represent the value set y can have.

follow that
y = x^2 +8, domain for x would be R, range for y would be y >=8
y =sqrt(5-x), domain for x would be x<=5, range for y would be y>=0

HTH