Questions: f(x)=x^2-8x+12 (a) The vertex is (4.0,-4.0). (Type an ordered pair, using integers or fractions.) (b) Determine whether the parabola has a maximum value or a minimum value and find the value. Select the correct choice below and fill in the answer box within your choice. (Type an integer or a fraction.) A. The parabola opens upward and has a minimum value of B. The parabola opens downward and has a maximum value of

Solution Steps

To solve the given problem, we need to analyze the quadratic function \( f(x) = x^2 - 8x + 12 \).

(a) The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Once we have the x-coordinate, we substitute it back into the function to find the y-coordinate of the vertex.

(b) To determine whether the parabola has a maximum or minimum value, we look at the coefficient of \( x^2 \) (which is \( a \)). If \( a > 0 \), the parabola opens upward and has a minimum value at the vertex. If \( a < 0 \), it opens downward and has a maximum value at the vertex. The value of the minimum or maximum is the y-coordinate of the vertex.

To find the vertex of the quadratic function \( f(x) = x^2 - 8x + 12 \), we use the formula for the x-coordinate of the vertex:

\[ x = -\frac{b}{2a} \]

Substituting \( a = 1 \) and \( b = -8 \):

\[ x = -\frac{-8}{2 \cdot 1} = 4.0 \]

Next, we substitute \( x = 4.0 \) back into the function to find the y-coordinate:

\[ f(4.0) = (4.0)^2 - 8 \cdot (4.0) + 12 = 16 - 32 + 12 = -4.0 \]

Thus, the vertex is \( (4.0, -4.0) \).

The coefficient of \( x^2 \) is \( a = 1 \), which is greater than zero. Therefore, the parabola opens upward, indicating that it has a minimum value at the vertex.

The minimum value of the function is the y-coordinate of the vertex, which is \( -4.0 \).

Final Answer