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 $-5$

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: Complete the square

Given (f(x)=2x^{2}+8x + 3). Factor out the coefficient of (x^{2}) from the first two terms: (f(x)=2(x^{2}+4x)+3). For the quadratic expression (x^{2}+4x), complete the square. ((x + 2)^{2}=x^{2}+4x+4), so (x^{2}+4x=(x + 2)^{2}-4). Substitute back: (f(x)=2((x + 2)^{2}-4)+3). Expand: (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 vertex form (y=a(x - h)^{2}+k), if (a>0), the parabola opens upward and has a minimum value at (x = h) with (y=k). Here (a = 2>0), so the function (f(x)=2(x + 2)^{2}-5) has a minimum value. The minimum value occurs when (x=-2), and (f(-2)=-5).

Answer:

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