Find the coordinates of the vertex for the parabola defined by the given quadratic function. \( f(x)=4 x^{2}-8 x+9 \)

The vertex of a parabola given by the quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex, which is \( x = -\frac{b}{2a} \). In this case, \( a = 4 \) and \( b = -8 \). Plugging in these values, we get \( x = -\frac{-8}{2 \cdot 4} = 1 \). To find the y-coordinate, substitute \( x = 1 \) back into the function: \( f(1) = 4(1)^2 - 8(1) + 9 = 5 \). Thus, the vertex is at the coordinates \((1, 5)\). If you want to deepen your understanding, try graphing the function! It visually reveals how the vertex forms the peak or trough, depending on whether the parabola opens upwards or downwards. You can also explore different values of \( a \) and see how they affect the shape and orientation of the parabola—it's like giving each curve a unique personality!