Answer :
To convert the vertex form of the equation [tex]\( y = \frac{1}{2}(x-4)^2 + 13 \)[/tex] to its standard form, follow these steps:
1. Start with the vertex form:
[tex]\[ y = \frac{1}{2}(x-4)^2 + 13 \][/tex]
2. Expand the squared term [tex]\( (x - 4)^2 \)[/tex]:
[tex]\[ (x-4)^2 = x^2 - 8x + 16 \][/tex]
3. Substitute this expanded form back into the original equation:
[tex]\[ y = \frac{1}{2}(x^2 - 8x + 16) + 13 \][/tex]
4. Distribute the [tex]\(\frac{1}{2}\)[/tex] across the terms inside the parentheses:
[tex]\[ y = \frac{1}{2}x^2 - \frac{1}{2}(8x) + \frac{1}{2}(16) + 13 \][/tex]
Simplifying each term:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 8 + 13 \][/tex]
5. Combine the constant terms:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 21 \][/tex]
This shows that the standard form of the equation is:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 21 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{D. \, y = \frac{1}{2} x^2 - 4 x + 21} \][/tex]
1. Start with the vertex form:
[tex]\[ y = \frac{1}{2}(x-4)^2 + 13 \][/tex]
2. Expand the squared term [tex]\( (x - 4)^2 \)[/tex]:
[tex]\[ (x-4)^2 = x^2 - 8x + 16 \][/tex]
3. Substitute this expanded form back into the original equation:
[tex]\[ y = \frac{1}{2}(x^2 - 8x + 16) + 13 \][/tex]
4. Distribute the [tex]\(\frac{1}{2}\)[/tex] across the terms inside the parentheses:
[tex]\[ y = \frac{1}{2}x^2 - \frac{1}{2}(8x) + \frac{1}{2}(16) + 13 \][/tex]
Simplifying each term:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 8 + 13 \][/tex]
5. Combine the constant terms:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 21 \][/tex]
This shows that the standard form of the equation is:
[tex]\[ y = \frac{1}{2}x^2 - 4x + 21 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{D. \, y = \frac{1}{2} x^2 - 4 x + 21} \][/tex]
