Question

Compare the functions shown below:
h(x)
f(x) = 2 sin (3x + Tt) - 2 g(x) = (x - 3)2 - 1
Which function has the smallest minimum y-value?

Answer 1

Answer:

The function with the smallest minimum y-value is h(x).

Step-by-step explanation:

f(x) = 2 sin (3x + π) - 2

The minimum value of the sine function is -1 so f(x) has a minimum value of

2 *-1 -2 = -4.

g(x) = (x - 3)^2 - 1

The minimum values of (x - 3)^2  is 0  so minimum for g(x) = -1.

From the table the minimum value for h(x) is -6.

Answer 2

Answer:

h had the lowest because -6 is less than both -4 and -1.

Step-by-step explanation:

The range of $y=\sin(x)$ is $-1\le y\le 1$.

The range of $y=2\sin(x)$ is $-2\le y \le 2$ since the amplitude has been changed from 1 or 2. (It has been vertically stretched.)

The range of $y=2\sin(3x+\pi)$ still had range $-2\le y \le 2$ because changing inside only effects the period and phase shift.

The range of $y=2\sin(3x+\pi)-2$ would have range $-2-2 \le y \le 2-2$ and after simplifying this you get the range is $-4\le y\le 0$.

The smallest value obtained by function f is -4.

$g(x)=(x-3)^2-1$ is a parabola in vertex form. The vertex is where the maximum or minimum of a parabola will occur.  It has a minimum since the coefficient of $x^2$ is positive.

Comparing to $a(x-h)^2+k$ where the vertex is (h,k) we should see that the vertex of g is at (3,-1). So the lowest y obtain by this parabola is -1.

So g lowest y is -1.

h is a list of points.  All you have to do is look through the second column to see which y is the lowest.

The lowest y there is -6 because -6 is less than all the other y's they have listed.

h has the lowest.

g has the highest.