Determine which coordinate is the vertex of \( f(x)=4 x^{2}-8 x+11 \) without graphing the parabola.

To find the vertex of the quadratic function \( f(x)=4 x^{2}-8 x+11 \), you can use the vertex formula. The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 4 \) and \( b = -8 \). Plugging in the values gives \( x = -\frac{-8}{2 \cdot 4} = \frac{8}{8} = 1 \). Now substitute \( x = 1 \) back into the function to find the y-coordinate: \[ f(1) = 4(1)^2 - 8(1) + 11 = 4 - 8 + 11 = 7 \] Thus, the vertex is at the point \( (1, 7) \). The vertex of a quadratic function is an important characteristic, as it represents either the maximum or minimum point of the graph. In this case, since the coefficient of \( x^2 \) is positive (i.e., \( a = 4 > 0 \)), the vertex at \( (1, 7) \) signifies the minimum point of the parabola. Knowing the vertex helps in understanding the overall shape of the graph without having to sketch it out!