Question

unit 03 classwork numbers: 1 - 9 score: 45 name: diego vilc date: september 04, 2025 pre - calculus period: 5 in exercises 1 - 3, determine whether the equation represents y as a function of x 1. x + y² = 4 2. |y| = 4 - x 3. (x + 3)² + y² = 1 in exercises 4 - 9, evaluate the function at each specified value of the independent variable and simplify. 4. h(t) = t² - 2t, h(2)=2² - 2(2)=4 - 4 h(2)=0

Explanation:

1. For the equation \(x + y^{2}=4\):

• # Explanation:

Step1: Isolate \(y^{2}\)

\(y^{2}=4 - x\)

Step2: Solve for \(y\)

\(y=\pm\sqrt{4 - x}\)
For a given \(x\) - value (e.g., when \(x = 0\), \(y=\pm2\)), there are two \(y\) - values. According to the definition of a function (for each input \(x\) in the domain, there is exactly one output \(y\)), \(y\) is not a function of \(x\).

• # Answer:

\(y\) is not a function of \(x\)

1. For the equation \(|y|=4 - x\):

• # Explanation:

Step1: Consider two cases

Case 1: When \(y\geq0\), \(y = 4 - x\). Case 2: When \(y\lt0\), \(y=x - 4\). For a given \(x\) - value (e.g., when \(x = 3\), if \(y\geq0\), \(y = 1\); if \(y\lt0\), \(y=-1\)), there are two \(y\) - values.

Step2: Apply function - definition

Since for some \(x\) values in the domain, there are two corresponding \(y\) values, \(y\) is not a function of \(x\) according to the definition of a function.

• # Answer:

\(y\) is not a function of \(x\)

1. For the equation \((x + 3)^{2}+y^{2}=1\):

• # Explanation:

Step1: Isolate \(y^{2}\)

\(y^{2}=1-(x + 3)^{2}\)

Step2: Solve for \(y\)

\(y=\pm\sqrt{1-(x + 3)^{2}}\). For a given \(x\) - value (e.g., when \(x=-3\), \(y=\pm1\)), there are two \(y\) - values.

Step3: Check function - condition

Since for some \(x\) values in the domain, there are two \(y\) values, \(y\) is not a function of \(x\) by the definition of a function.

• # Answer:

\(y\) is not a function of \(x\)

1. For the function \(h(t)=t^{2}-2t\), evaluating \(h(2)\):

• # Explanation:

Step1: Substitute \(t = 2\) into the function

\(h(2)=2^{2}-2\times2\)

Step2: Simplify the expression

\(h(2)=4 - 4=0\)

• # Answer:

\(h(2)=0\)

Answer:

1. For the equation \(x + y^{2}=4\):

• # Explanation:

Step1: Isolate \(y^{2}\)

\(y^{2}=4 - x\)

Step2: Solve for \(y\)

\(y=\pm\sqrt{4 - x}\)
For a given \(x\) - value (e.g., when \(x = 0\), \(y=\pm2\)), there are two \(y\) - values. According to the definition of a function (for each input \(x\) in the domain, there is exactly one output \(y\)), \(y\) is not a function of \(x\).

• # Answer:

\(y\) is not a function of \(x\)

1. For the equation \(|y|=4 - x\):

• # Explanation:

Step1: Consider two cases

Case 1: When \(y\geq0\), \(y = 4 - x\). Case 2: When \(y\lt0\), \(y=x - 4\). For a given \(x\) - value (e.g., when \(x = 3\), if \(y\geq0\), \(y = 1\); if \(y\lt0\), \(y=-1\)), there are two \(y\) - values.

Step2: Apply function - definition

Since for some \(x\) values in the domain, there are two corresponding \(y\) values, \(y\) is not a function of \(x\) according to the definition of a function.

• # Answer:

\(y\) is not a function of \(x\)

1. For the equation \((x + 3)^{2}+y^{2}=1\):

• # Explanation:

Step1: Isolate \(y^{2}\)

\(y^{2}=1-(x + 3)^{2}\)

Step2: Solve for \(y\)

\(y=\pm\sqrt{1-(x + 3)^{2}}\). For a given \(x\) - value (e.g., when \(x=-3\), \(y=\pm1\)), there are two \(y\) - values.

Step3: Check function - condition

Since for some \(x\) values in the domain, there are two \(y\) values, \(y\) is not a function of \(x\) by the definition of a function.

• # Answer:

\(y\) is not a function of \(x\)

1. For the function \(h(t)=t^{2}-2t\), evaluating \(h(2)\):

• # Explanation:

Step1: Substitute \(t = 2\) into the function

\(h(2)=2^{2}-2\times2\)

Step2: Simplify the expression

\(h(2)=4 - 4=0\)

• # Answer:

\(h(2)=0\)