Which of the following quadratics will have a minimum value of -5 at \( x=7 \) ? \( \begin{array}{ll}\text { (1) } y=x^{2}-14 x+39 & \text { (3) } y=x^{2}-14 x+44 \\ \text { (2) } y=-x^{2}+14 x-54 & \text { (4) } y=-x^{2}-10 x-18\end{array} \)

To find out which quadratic has a minimum value of -5 at \( x=7 \), we can use the vertex form of a parabola. The vertex occurs at \( x = -\frac{b}{2a} \). For cases (1) and (3), we find that they have a positive leading coefficient \( a=1 \), meaning they open upwards and their vertex gives the minimum point. In this case, the x-value at the vertex can be calculated using the equation and should equal 7. Calculating for option (1): \[y = x^2 - 14x + 39\] The vertex occurs at \( x = 7 \). Checking for minimum value: \[ y(7) = (7)^2 - 14(7) + 39 = 49 - 98 + 39 = -10 \quad \text{(not -5)} \] Calculating for option (2): \[y = -x^2 + 14x - 54\] The vertex occurs at \( x = 7 \). Checking the value: \[ y(7) = - (7)^2 + 14(7) - 54 = -49 + 98 - 54 = -5 \quad \text{(yes!)} \] Option (3) gives: \[y = x^2 - 14x + 44 \] Calculating: \[ y(7) = (7)^2 - 14(7) + 44 = 49 - 98 + 44 = -5 \quad \text{(also works!)} \] Both options (2) and (3) give a minimum value of -5 at \( x = 7 \). So, the answer is: **Quadratic (2) and Quadratic (3)**.