Question

What is the domain and range of the quadratic equation y = (x – 5)^2+ 10?

Answer 1

y=(x-5)^(2)+10

The domain of the rational expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
All real numbers

Answer 2

y=(x-5)^(2)+10

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
(x-5)^(2)+10=y

Since 10 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 10 from both sides.
(x-5)^(2)=-10+y

Move all terms not containing x to the right-hand side of the equation.
(x-5)^(2)=y-10

Take the square root of each side of the equation to setup the solution for x.
~((x-5)^(2))=\~(y-10)

Remove the perfect root factor (x-5) under the radical to solve for x.
(x-5)=\~(y-10)

First, substitute in the + portion of the \ to find the first solution.
(x-5)=~(y-10)

Remove the parentheses around the expression x-5.
x-5=~(y-10)

Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides.
x=5+~(y-10)

Next, substitute in the - portion of the \ to find the second solution.
(x-5)=-~(y-10)

Remove the parentheses around the expression x-5.
x-5=-~(y-10)

Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides.
x=5-~(y-10)

The complete solution is the result of both the + and - portions of the solution.
x=5+~(y-10),5-~(y-10)

The domain of an expression is all real numbers except for the regions where the expression is undefined. This can occur where the denominator equals 0, a square root is less than 0, or a logarithm is less than or equal to 0. All of these are undefined and therefore are not part of the domain.
(y-10)<0

Solve the equation to find where the original expression is undefined.
y<10

The domain of the rational expression is all real numbers except where the expression is undefined.
y>=10_[10,<Z>I<z>)

The domain of the inverse of y=(x-5)^(2)+10 is equal to the range of f(y)=5+~(y-10).
Range: y>=10_[10,<Z>I<z>)

Answer 3

?

Answer 4

domain of a polynomial is all real numbers.

Answer 5

range of
\[y=(x-5)^2=10\]
is
\[[10,\infty)\] with your eyeballs since
\[(x-5)^2 \geq 0\] since it is a square. and since it is greater than or equal to zero, the whole thing is greater than or equal to 10

Answer 6

don't be moving stuff left and right.
\[y=(x-5)^2+10\] is a parabola facing up with

1) vertex (5,10) so
2) minimum value 10
3) range greater than or equal to 10

all three statements say basically the same thing

Answer 7

Minimum value? Value of y, right? If x is 0 then 25+10 = 35

Answer 8

i got this one guys haha move to y next questions plz :)

Answer 9

hold on
minimum values does mean minimum value of y correct. but in this case that occurs where x = 5 not x = 0

Answer 10

Aha. OK.

Answer 11

it is clear yes?
\[(x-5)^2=0\] means x = 5

Answer 12

yes.