y = -x2 + 1
This function describes a parabola that opens downward. To find the top of it's range, you need to find it's focal point. You can do that very easily by taking the derivative of the equation and solving it for 0:
y = -x2 + 1
∴ y' = -2x
let y' = 0:
0 = -2x
∴ x = 0
Now you can calculate the y value at that point:
y = -02 + 1
∴ y = 1
So that function describes an upside down parabola whose peak is at the point {0, 1}. It's range then is:
{y | y ∈ ℜ, y ≤ 1}
The answer depends on the domain. If the domain is the whole of the real numbers, the range in y ≥ 1. However, you can choose to have the domain as [1, 2] in which case the range will be [2, 5]. If you choose another domain you will get another range.
if x2 + 7 = 37, then x2 = 29 and x = ±√29
y = x2 + 2x + 1zeros are:0 = x2 + 2x + 10 = (x + 1)(x + 1)0 = (x + 1)2x = -1So that the graph of the function y = x + 2x + 1 touches the x-axis at x = -1.
The discriminant formula. b2 - 4ac 32 - 4(1)(8) 9 - 32 = - 23 ===========This shows no real roots to this function.
f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28
Yes
No, it is not.
y ≥ 11
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
x2 + x2 = 2x2
Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2
(the shape is an upside down 'u').
x = -3y = -14
x2+2x+1=y or y=x2 In this function the domain is x equals real values and the range is y equals all real values provided y is more than or equal to zero.
The answer depends on the domain. If the domain is the whole of the real numbers, the range in y ≥ 1. However, you can choose to have the domain as [1, 2] in which case the range will be [2, 5]. If you choose another domain you will get another range.
It equals x2 - 34
It is a many to one function (APEX)