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f(x) = x2 + 7x + 9 (a) Write f(x) in the form k(x + t)2 + r. Fill in the obtained values of k, t, r: k = t = r = (b) Find the value of x where f(x) attains its minimum value or its maximum value. x =
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x^2+7x+9 \[(x+\frac{7}{2})^2-3.25\]so, k=1 t=3.5 r=3.25
global min occurs at x=-7/2
f(x)=x^2 + 7x +9 by completing the sq, f(x) = (x+7/2)^2 - (7/2)^2 +9 f(x) = (x+7/2)^2 - (13/4) k=1, t=7/2, r=-13/4 since k=1, f(x) has a min value at x+(7/2)=0 x=-7/2 when y is at -13/4, your vertex is (-7/2 , -13/4)
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