1. What are the domain and range of g(x) = -1/4 (x−17)2 + 61

Question

1. What are the domain and range of g(x) = -1/4 (x−17)² + 61

A. Domain: All real numbers

Range: g(x) ≤ 61

B. Domain: x ≤ 17

Range: g(x) ≤ 61

C. Domain: All real numbers

Range: x ≤ 17

D. Domain: g(x) ≥ 61

Range: x ≤ 17

Answer 1

The Vertex Form of a Parabola is:

g(x) = a(x - h)² + k

vertex = (h, k)

rate = a (positive is up; negative is down)

a = -1/4 (downward parabola)

vertex = (17, 61)

There is no domain restriction.

Since the parabola is upside down (downward-facing), it reaches it's maximum at its vertex. In other words, is not possible for g(x) to be greater than 61.

Answer 2

Domain is the set of all possible x values that can be plugged into a function and yield a real number. If an x value causes a denominator of 0 that value would not be part of the domain. If the number causes a square root of a negative number, it also would not be part of the domain.

Since all x values that are could be plugged in yield a real number (there would be no way to get a denominator of 0 or square root of a negative), all real numbers are going to be in the domain.

Let's consider numbers that we get for our outputs when looking for the range.

Range is the set of all possible y values that can "come out" of your function. So if you are looking at domain you are trying to figure out what are the biggest and/or smallest numbers that can come out of this function. Are their limits to the y values I get when I plug in values from my domain?

0 is usually an easy baseline number to start with. Let experiment and let's consider what happens when you plug in 0 for x.

g(0) = -1/4 (0−17)² + 61 = -1/4(-17)²+61 =-1/4(-17*-17)+61 = -1/4(289)+61 = -72.25+61=-11.25

Can we find a number smaller than this? Bigger than this? To explore possible range values, let's try to consider what x value we can plug in to (x-17)² to make it as small as possible and see what happens to the rest of the equation. Well the smallest number we can make here is 0. Since (x-17)² will never be able to produce negative numbers it's smallest number we can get out of (x-17)² has to be 0. x=17 makes (x-17)² = 0 so let's see what happens to the overall function when we plug in 17 to the overall function.

g(17) = -1/4 (17−17)² + 61 = -1/4 (0*0)+61= 0+61= 61

So...

1. When we plugged in 0 we ended up with a number smaller than 61 because -1/4(0-17)² yielded a negative number. And a negative number+ 61 must be smaller than 61.
2. When we plugged in 17 we ended up with a number equal to 61 because -1/4(17-17)² yielded 0. And 0+61 must equal 61.

The question now is is there a way to get a number bigger than 61?

Well we would need a positive number+61 to accomplish this. Which means we would need to make -1/4(x-17)² equal a positive number. To do this (x-17)² would have to yield a negative number. That way -1/4 would be multiplied by a negative number and become positive.

Any negative number squared equals a positive number. Any positive number squared is a positive number. SO the is NO way to make -1/4(x-17)² yield a positive number.

This means I will never be able to have the overall function (g(x) = -1/4 (x−17)² + 61) equal a number bigger than 61 because -1/4(x-17)² will never be a positive number.

Conclusion: the function can yield y values/outputs smaller than 61 and equal to 61, but it cannot yield numbers bigger than 61. SO the range is all numbers equal to or smaller than 61. (g(x)≤61)

This means that A must be the correct answer.