It is in quadrants 1 and 2
It is v shaped
it goes through the origin
hope this helps!
It’s vertex is not at the origin
Its vertex is not at the origin
It is a reflection of the original graph in the line y = x.
The parent function of the exponential function is ax
Reciprocal parent function
apex what is the range of the absolute... answer is nonnegative real num...
the range is all real numbers
F(x)=x
apex what is the range of the absolute... answer is nonnegative real num...
It’s vertex is not at the origin
It is a hyperbola, it is in quadrants I and II
The graph of the absolute value parent function, ( f(x) = |x| ), has a V-shape with its vertex at the origin (0, 0). It is symmetric about the y-axis, indicating that it is an even function. The graph consists of two linear segments that extend infinitely in the positive y-direction, with a slope of 1 for ( x \geq 0 ) and a slope of -1 for ( x < 0 ). Additionally, it never dips below the x-axis, as absolute values are always non-negative.
The domain of the absolute value parent function, ( f(x) = |x| ), is all real numbers, expressed as ( (-\infty, \infty) ). The range is all non-negative real numbers, represented as ( [0, \infty) ), since the absolute value cannot be negative.
Its vertex is not at the origin
i believe it is a linear linegoing diagonally
Please don't write "the following" if you don't provide a list.
f(x) = |f(x)|/3