if y = x2 + 1 Then the minimum value of y is 1, which happens at the point (0, 1). It lies in the domain of real numbers. i.e. {y | y ≥ 1, y ∈ ℝ}
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.
4
y=(√1)-x2 The domain is the set of numbers that "x" can be. In this equation "x" can be any real number. The domain for this problem would be (-inf,inf) *Inf= Infinity*
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
if y = x2 + 1 Then the minimum value of y is 1, which happens at the point (0, 1). It lies in the domain of real numbers. i.e. {y | y ≥ 1, y ∈ ℝ}
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.
If the domain is the set of reals, then the range is the whole set of non-negative reals.
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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 you mean y = x2, then yes, it is nonlinear.
y=(√1)-x2 The domain is the set of numbers that "x" can be. In this equation "x" can be any real number. The domain for this problem would be (-inf,inf) *Inf= Infinity*
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
The domain of y = 1/x2 is all numbers from -infinity to + infinity except zero. The range is all positive numbers from zero to +infinity, except +infinity.
No translation will invert a quadratic graph.
The domain of y = -4x is all real numbers.
the graph is moved down 6 units