SOLUTION: What are the x and y intercepts of these equations 1. f(x)={√x^2, -3≤ x< 0 {x^2 , 0≤ x ≥ 2 2. f(x)={2x, -3≤ x< 0 {1, 0≤ x ≥ 4 3. f(x)={x, 0≤ x< 5 {x-1, -2≤ x

Algebra ->  Graphs -> SOLUTION: What are the x and y intercepts of these equations 1. f(x)={√x^2, -3≤ x< 0 {x^2 , 0≤ x ≥ 2 2. f(x)={2x, -3≤ x< 0 {1, 0≤ x ≥ 4 3. f(x)={x, 0≤ x< 5 {x-1, -2≤ x       Log On


   



Question 1153427: What are the x and y intercepts of these equations
1. f(x)={√x^2, -3≤ x< 0 {x^2 , 0≤ x ≥ 2
2. f(x)={2x, -3≤ x< 0 {1, 0≤ x ≥ 4
3. f(x)={x, 0≤ x< 5 {x-1, -2≤ x ≥ 2

Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


Note that none of the descriptions of the functions is valid; specifying a range of values as "0≤ x ≥ 2" is meaningless. I will assume that in each example the intended range is like "0≤ x ≤ 2"
1. y+=+sqrt%28x%5E2%29 on [-3,0);
   y+=+x%5E2 on [0,2]

For negative values of x, sqrt%28x%5E2%29+=+-x. On the interval [-3,0), neither x nor -x has the value 0; there are no x- or y-intercepts on that part of the
graph.

For non-negative values of x, on the interval [0,2] both the x and y values are 0 only at x=0. The x- and y-intercepts are both (0,0).

2. y+=+2x on [-3,0);
   y+=+1 on [0,4]

On the interval [-3,0), x is always negative, so y=2x is always negative; there are no x- or y-intercepts on that part of the graph.

On the interval [0,4], x is clearly 0 at x=0; y is always 1, so it is never 0. This function has an x-intercept at (0,1) and no y-intercept.

3. y+=+x on [0,5);
   y+=+x-1 on [-2,2]

This definition is not valid; the definition results in two different values on the interval [0,2].