Question

determine the vertex form and the maximum or minimum value of the function. f(x)=2x²+8x+3 a f(x)=2(x+4)²−13, the maximum value of f is −13 b f(x)=2(x+4)²−13, the minimum value of f is −13 c f(x)=2(x+2)²−5, the maximum value of f is −5 d f(x)=2(x+2)²−5, the minimum value of f is −5

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 \(x^{2}+4x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a = x\) and \(2ab=4x\) (so \(b = 2\)), then \(x^{2}+4x=(x + 2)^{2}-4\).
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 \(y=a(x - h)^{2}+k\), when \(a>0\), the parabola opens upward and the vertex \((h,k)\) is a minimum point.
Here \(a = 2>0\), \(h=-2\), \(k=-5\).

Answer:

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