f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28
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.
x = -5 x = 2
x2+11x+11 = 7x+9 x2+11x-7x+11-9 = 0 x2+4x+2 = 0 The above quadratic equation can be solved by using the quadratic equation formula and it will have two solutions.
No, it is not.
Yes
No. By definition, a function has a single unique value for every value passed into it. The equation given here describes a circle, which can not be rearranged to meet this condition.
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
We can't calculate what it equals until we know the value of ' x '.
x2 + x2 = 2x2
x=-11
Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2
8
You need to know the value of x
x = -3y = -14
(the shape is an upside down 'u').