Question

cc.2.2.hs.d.7
determine whether $f(x) = 4x^2 - 16x + 6$ has a maximum or a minimum value and find that value.
\bigcirc minimum; $(2, -10)$
\bigcirc minimum; $(-10, 2)$
\bigcirc maximum; $(2, -10)$
\bigcirc maximum; $(-10, 2)$

Explanation:

Step1: Analyze the coefficient of \(x^2\)

For a quadratic function \(f(x)=ax^2 + bx + c\), if \(a>0\), the parabola opens upward, so the function has a minimum value. Here, \(a = 4>0\), so \(f(x)\) has a minimum.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of \(f(x)=ax^2+bx + c\) is given by \(x=-\frac{b}{2a}\). For \(f(x)=4x^2-16x + 6\), \(a = 4\) and \(b=-16\). Then \(x=-\frac{-16}{2\times4}=\frac{16}{8} = 2\).

Step3: Find the y - coordinate of the vertex

Substitute \(x = 2\) into the function \(f(x)=4x^2-16x + 6\). \(f(2)=4\times(2)^2-16\times2 + 6=4\times4-32 + 6=16-32 + 6=-10\). So the vertex is \((2,-10)\), which is the point where the minimum value occurs.

Answer:

A. minimum; (2, - 10)