1. a) Express \( y=x^{2}+8 x+9 \) in the form \( y=a(x-h)^{2}+k \) b) For what value of \( x \) is \( y=x^{2}+8 x+9 \) a maximum? \( y=1(x-8)^{2}+9 \)

To express \( y = x^2 + 8x + 9 \) in vertex form, we can complete the square. Start with \( x^2 + 8x \); take half of 8 (which is 4), square it (resulting in 16), and you can rewrite the equation as \( y = (x + 4)^2 - 16 + 9 \). Thus, we get \( y = (x + 4)^2 - 7 \), so \( a = 1, h = -4, k = -7 \). Since the coefficient of \( x^2 \) is positive, the parabola opens upwards, and there’s no maximum value. However, the minimum value occurs at the vertex \( x = -4 \), producing the minimum value of \( y = -7 \). Hence, \( y \) doesn’t have a maximum value, but rather a minimum where \( x = -4 \) and \( y = -7 \).