Question

etermine the vertex form and the maximum or minimum value of the function.\\( f(x) = 2x^2 + 8x + 3 \\)\\( \boldsymbol{a} \\) \\( f(x) = 2(x + 4)^2 - 13 \\), the maximum value of \\( f \\) is \\( -13 \\)\\( \boldsymbol{b} \\) \\( f(x) = 2(x + 4)^2 - 13 \\), the minimum value of \\( f \\) is \\( -13 \\)\\( \boldsymbol{c} \\) \\( f(x) = 2(x + 2)^2 - 5 \\), the maximum value of \\( f \\) is \\( -5 \\)\\( \boldsymbol{d} \\) \\( f(x) = 2(x + 2)^2 - 5 \\), the minimum value of \\( f \\) is \\( -5 \\)

Explanation:

Step1: Complete the square for the quadratic function

Given \( f(x) = 2x^2 + 8x + 3 \), first factor out the coefficient of \( x^2 \) from the first two terms:
\( f(x) = 2(x^2 + 4x) + 3 \)

To complete the square inside the parentheses, take half of the coefficient of \( x \) (which is \( 4 \)), square it: \( (\frac{4}{2})^2 = 4 \). Add and subtract this value inside the parentheses:
\( f(x) = 2(x^2 + 4x + 4 - 4) + 3 \)

Rewrite the expression:
\( f(x) = 2((x + 2)^2 - 4) + 3 \)

Distribute the \( 2 \):
\( f(x) = 2(x + 2)^2 - 8 + 3 \)

Simplify:
\( f(x) = 2(x + 2)^2 - 5 \)

Step2: Determine maximum or minimum value

For a quadratic function in the form \( f(x) = a(x - h)^2 + k \), if \( a > 0 \), the parabola opens upward, so the function has a minimum value at \( k \). Here, \( a = 2 > 0 \), so the function has a minimum value of \( -5 \) (since \( k = -5 \)).

• By completing the square, we convert \( f(x) = 2x^2 + 8x + 3 \) to vertex form \( f(x) = 2(x + 2)^2 - 5 \).
• Since the coefficient of \( x^2 \) ( \( a = 2 \)) is positive, the parabola opens upward, meaning the function has a minimum value at the vertex's \( y \)-coordinate (\( k = -5 \)).

Answer:

D. \( f(x) = 2(x + 2)^2 - 5 \), the minimum value of \( f \) is \( -5 \)