determine the vertex form and the maximum or minimum value of the function. \n$f(x) = 2x^2 + 8x + 3$\na…

determine the vertex form and the maximum or minimum value of the function. \n$f(x) = 2x^2 + 8x + 3$\na $f(x) = 2(x + 4)^2 - 13$, the maximum value of $f$ is $-13$\nb $f(x) = 2(x + 4)^2 - 13$, the minimum value of $f$ is $-13$\nc $f(x) = 2(x + 2)^2 - 5$, the maximum value of $f$ is $-5$\nd $f(x) = 2(x + 2)^2 - 5$, the minimum value of $f$ is $-5$
Answer
Explanation:
Step1: Recall Vertex Form
The vertex form of a quadratic function is ( f(x) = a(x - h)^2 + k ), where ((h, k)) is the vertex. For ( f(x) = 2x^2 + 8x + 3 ), we complete the square.
Step2: Factor Out Coefficient of ( x^2 )
Factor out 2 from the first two terms: ( f(x) = 2(x^2 + 4x) + 3 ).
Step3: Complete the Square Inside the Parentheses
Take half of 4 (which is 2), square it (4), and add/subtract inside: ( f(x) = 2(x^2 + 4x + 4 - 4) + 3 ).
Step4: Rewrite as Perfect Square
Simplify: ( f(x) = 2((x + 2)^2 - 4) + 3 = 2(x + 2)^2 - 8 + 3 = 2(x + 2)^2 - 5 ).
Step5: Determine Max/Min
Since ( a = 2 > 0 ), the parabola opens upward, so there's a minimum at ( k = -5 ).
Answer:
D. ( f(x) = 2(x + 2)^2 - 5 ), the minimum value of ( f ) is ( -5 )