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 for the quadratic function

Given $f(x)=2x^{2}+8x + 3$. First, factor out the coefficient of $x^{2}$ from the first - two terms: $f(x)=2(x^{2}+4x)+3$. For the expression inside the parentheses $x^{2}+4x$, we know that $(a + b)^2=a^{2}+2ab + b^{2}$. Here $a = x$ and $2ab = 4x$, so $b = 2$ and $x^{2}+4x=(x + 2)^{2}-4$. Then $f(x)=2((x + 2)^{2}-4)+3$.

Step2: Expand the expression

Expand $2((x + 2)^{2}-4)+3$: [ \begin{align*} f(x)&=2(x + 2)^{2}-8 + 3\ &=2(x + 2)^{2}-5 \end{align*} ] Since the coefficient of $(x + 2)^{2}$ is $a = 2>0$, the parabola opens upward and the function has a minimum value. The vertex - form of a quadratic function is $y=a(x - h)^{2}+k$, and the vertex is $(h,k)$. For $y = 2(x + 2)^{2}-5$, the vertex is $(-2,-5)$ and the minimum value of the function is $-5$.

Answer:

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