Question

determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find the value.

f(x) = 3x^2 - 12x

the quadratic function has a value.

Explanation:

Step1: Identify coefficients

For the quadratic function $f(x)=3x^{2}-12x$, $a = 3$, $b=- 12$, $c = 0$.

Step2: Determine max - min

Since $a = 3>0$, the parabola opens upward, so the function has a minimum value.

Step3: Find x - coordinate of vertex

The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a = 3$ and $b=-12$ into the formula: $x=-\frac{-12}{2\times3}=\frac{12}{6} = 2$.

Step4: Find the minimum value

Substitute $x = 2$ into the function $f(x)=3x^{2}-12x$. Then $f(2)=3\times2^{2}-12\times2=3\times4 - 24=12 - 24=-12$.

Answer:

The quadratic function has a minimum value of - 12.