If r(x) = 2 – x2 and w(x) = x – 2, what is the range of (wºr) (x)
(-∞,0]
(-∞,2]
[0,∞)
[2,∞)

(-inf,2]

Step-by-step explanation:

I replaced r(x) with 2-x^2

I replace the x in w(x)=x-2 with 2-x^2

You can graph this to find the range.

But since this is a quadratic (the graph is a parabola), I'm going to find the vertex to help me to determine the range.

The vertex is at x=-b/(2a).  Once I find x, I can find the y that corresponds to it by using y=-x^2-x+2.

Comparing ax^2+bx+c to -x^2-x+2 tells me a=-1, b=-1, and c=2.

So the vertex is at x=1/(2*-1)=-1/2.

To find the y-coordinate that corresponds to that I will not plug in -1/2 in place of x into -x^2-x+2.

This gives me

-(-1/2)^2-(-1/2)+2

-1/4 +  1/2  +2

Find a common denominator which is 4.

-1/4 +  2/4  +8/4

8/4

2.

So the highest y value is 2 ( I know tha parabola is upside down because a=negative number)

That mean then range is 2 or less than 2.

So the answer an interval notation is (-inf,2]

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.