Question

7.
a
b

1. identify the minimum or maximum value and the domain and range of the graphs of the functions

y = (x + 3)^2 - 4
a. minimum value: 4
domain: all real numbers ≥ 4
range: all real numbers
b. minimum value: -4
domain: all real numbers
range: all real numbers ≥ -4
what is the relative maximum and minimum of the function?

1. h(x) = 3x^3 + x^2 - 3x

a. the relative maximum is at (-0.85, 1.61) and the
relative minimum is at (0.60, -1.61)
b. the relative minimum is at (-0.85, 1.61) and the
relative maximum is at (0.60, -1.61)

1. 10. classify -x^3 + 3x^2 - 1x^4 by degree

a. quartic
b. quartic

1. 11.

\

$$\begin{bmatrix} 3 & -4 & 1 \\\\ 0 & 3 & 2 \\\\ \\end{bmatrix}$$

+ \

$$\begin{bmatrix} -5 & 0 & 3 \\\\ -4 & 3 & -1 \\\\ \\end{bmatrix}$$

a. \

$$\begin{bmatrix} -2 & -4 & 0 \\\\ -4 & 3 & 1 \\\\ \\end{bmatrix}$$

b. \

$$\begin{bmatrix} -2 & -4 & 4 \\\\ -4 & 6 & 1 \\\\ \\end{bmatrix}$$

Explanation:

Step1: Analyze the Function Type

The function given is \( y=(x + 2)^2-4 \), which is a quadratic function in vertex form \( y=a(x - h)^2+k \), where \( a = 1\), \( h=-2\), \( k = - 4\). For a quadratic function \( y=a(x - h)^2+k \), if \( a>0\), the parabola opens upwards, so the vertex \((h,k)\) is the minimum point. Here \( a = 1>0\), so the function has a minimum value at \( k=-4\)? Wait, no, wait: the vertex form is \( y=a(x - h)^2 + k\), so when \( a = 1\), \( h=-2\), \( k=-4\), so the vertex is \((-2,-4)\) and since \( a>0\), the parabola opens upwards, so the minimum value is \( k=-4\)? Wait, no, the equation is \( y=(x + 2)^2-4\), which can be written as \( y=(x-(-2))^2+(-4)\). So the vertex is \((-2,-4)\) and since \( a = 1>0\), the parabola opens upward, so the minimum value is \( y=-4\) (when \( x=-2\)). The domain of a quadratic function is all real numbers because there are no restrictions on \( x\) (we can plug in any real number for \( x\)). The range: since the parabola opens upward and the minimum value is \( y=-4\), the range is all real numbers greater than or equal to \(-4\)? Wait, no, wait the options: let's check the options. The options are:

a. minimum value: 4; domain: all real numbers ≥4; range: all real numbers

b. minimum value: -4; domain: all real numbers; range: all real numbers ≥ -4

Wait, maybe I misread the original function. Wait the original problem says "Identify the minimum or maximum value and the domain and range of the graphs of the function \( y=(x + 2)^2-4 \)".

So let's re-express:

For the function \( y=(x + 2)^2-4 \):

• Since the coefficient of \((x + 2)^2\) is \( 1>0\), the parabola opens upward, so it has a minimum value.

• The vertex of the parabola is at \((h,k)\) where \( h=-2\), \( k=-4\), so the minimum value is \( k=-4\) (when \( x=-2\)).

• Domain: For any quadratic function \( y = ax^2+bx + c\) (or in vertex form), the domain is all real numbers because there's no value of \( x\) that makes the function undefined (we can square any real number, add, subtract, etc.). So domain is all real numbers.

• Range: Since the parabola opens upward and the minimum \( y\)-value is \(-4\), the range is all real numbers \( y\) such that \( y\geq - 4\) (because the parabola goes from \( y=-4\) upwards to infinity).

Now let's check the options:

Option a: minimum value: 4; domain: all real numbers ≥4; range: all real numbers → This is incorrect because the minimum value is -4, domain is all real numbers, range is \( y\geq -4\).

Option b: minimum value: -4; domain: all real numbers; range: all real numbers ≥ -4 → This matches our analysis.

Wait, but the original problem's options might have been miswritten? Wait the user's image shows:

"8. Identify the minimum or maximum value and the domain and range of the graphs of the function \( y=(x + 2)^2-4 \)

a. minimum value: 4

domain: all real numbers ≥4

range: all real numbers

b. minimum value: -4

domain: all real numbers

range: all real numbers ≥ -4"

So the correct option is b.

Step2: Confirm the Analysis

• The function is a quadratic with \( a = 1>0\), so it opens upward, hence has a minimum at the vertex.

• The vertex is at \( x=-2\), \( y=-4\), so minimum value is -4.

• Domain: All real numbers (no restrictions on \( x\)).

• Range: Since the parabola opens upward and the minimum \( y\) is -4, range is \( y\geq -4\).

So option b is correct.

Answer:

b. minimum value: -4; domain: all real numbers; range: all real numbers ≥ -4