sketch the graph of the following function, i. y=6-square root of -x. ii. y=(x-1)square root of 2-16.

Question

1. sketch the graph of the following function, i. y=6-square root of -x. ii. y=(x-1)square root of 2-16.

label graph and indicate the coordinate of the turning points, corner points, x  and y intercept. Determine the symmetry of each graph, determine the intervals for x where why is increasing, decreasing or constant and for each problem solve the inequalities y is less then equal to 0 and y is greater than 0. find the domain and range of each function and plot the graph of absolute value of y.  

Answer 1

i)

 

y = 6 - √(-x)

 

 

First, we need to find the domain before we even begin to graph this.  Since we cannot have a negative number under the square-root, the domain is all numbers less than or equal to 0.  In interval notation, 

 

(-∞, 0].

 

Here is why.

 

The number under the square-root of this function will then be -(-x) = x.  This is a positive number under the square-root, which is acceptable.

 

To find the x-intercept, we set y=0 and solve for x.

 

0 = 6 - √(-x)

 

-6 = - √(-x)

 

6 = √(-x)

 

36 = -x

 

-36 = x

 

The x-intercept is (-36, 0).

 

The y-intercept is (0, 6).

 

 

You will plot these two points on a coordinate system.  In addition to them, plot these points at x =  -1 , -4 , -9 , -16 , -25 and so on.  Notice that I picked numbers so that when you take the square-root of them, the result is an integer.  It makes it easier to plot.  The negative sign only indicates the domain.  Once you plot these points and connect them, it will be easier to find the intervals of increase and decrease, turning points, corner points, and range.

 

Other stuff to know:

 

Turning points are single points where the graph starts to turn or change direction.

Corner points are single points where the graph intersects a region in the coordinate system.

Range is the set of y values in which the function exists.

 

 

Do the same for the second problem.