the function f is defined:
f(x)= x+3 if -2<=x<1
9 if x=1
-x+2 if x>1
a. find the domain
b. locate the intercepts
I am totally lost, can someone please help me with this?
f(x0 = x+3 for -2 ≤ x < 1
f(x) = 9 , for x = 1
f(x) = -x+2 , for x > 1
strange graph.
I see a line segment from (-2,1) to (1,4) with a solid dot at (-2,1) (include the point) and an open dot at (1,4), exclude the point
The we have the isolated and lonely point (1,9) just sitting there all by itself.
then a ray starting at (1,1) (open dot, point not included)
with an x-intercept at (2,0) and dropping down into the fourth quadrant.
domain: x≥ -2
intercepts,
y-intercept at (0,3)
x-intercept at (2,0)
Of course! I'll help you step by step.
a. To find the domain of a function, you need to determine the set of all possible input values (or x-values) for which the function is defined. In this case, we have three different cases for the function f(x) depending on the value of x.
1. When -2 <= x < 1, the function f(x) is defined as f(x) = x + 3.
2. When x = 1, the function f(x) is defined as f(x) = 9.
3. When x > 1, the function f(x) is defined as f(x) = -x + 2.
Since there are no restrictions or limitations on the input values for any of these cases (except for x <= 1 in the first case), the domain of the function f(x) is all real numbers, (-∞, +∞).
b. To locate the intercepts, we need to find where the function intersects the x-axis (x-intercepts) and where it intersects the y-axis (y-intercepts).
1. x-intercepts: These are the points on the graph where the function intersects the x-axis, meaning the y-coordinate is zero. To find the x-intercept, set f(x) = 0 and solve for x.
In the first case (-2 <= x < 1), we have f(x) = x + 3. Setting f(x) = 0, we get:
x + 3 = 0
x = -3
So, the x-intercept is at (-3, 0).
2. y-intercept: This is the point on the graph where the function intersects the y-axis, meaning the x-coordinate is zero. To find the y-intercept, substitute x = 0 into the function f(x).
In the first case (-2 <= x < 1), we have f(x) = x + 3. Substituting x = 0, we get:
f(0) = 0 + 3
f(0) = 3
So, the y-intercept is at (0, 3).
Since the other cases when x = 1 or x > 1 have specific function values, there are no additional intercepts.
I hope this helps! Let me know if you have any further questions.