answer:- **Step 1: Find the direction vectors** [ vec{v_1} = begin{pmatrix} 2 -3 end{pmatrix} - begin{pmatrix} -3 2 end{pmatrix} = begin{pmatrix} 5 -5 end{pmatrix} ] [ vec{v_2} = begin{pmatrix} 4 -5 end{pmatrix} - begin{pmatrix} -3 2 end{pmatrix} = begin{pmatrix} 7 -7 end{pmatrix} ] - **Step 2: Checking for collinearity** The vectors vec{v_1} and vec{v_2} should be proportional. We see that: [ frac{7}{5} = frac{-7}{-5} = 1.4 ] This confirms that both vec{v_1} and vec{v_2} are proportional, thus confirming collinearity. - **Step 3: Normalize the direction vector** We choose to use vec{v_1} for simplicity: [ text{Normalized form} = frac{1}{5}begin{pmatrix} 5 -5 end{pmatrix} = begin{pmatrix} 1 -1 end{pmatrix} ] Thus, b = boxed{1}. Conclusion: Factorization shows that all three points are collinear and the desired parameterized vector form is achieved with b = 1.

answer:First, note that the vertex form of a parabola's equation is y = a(x-h)^2 + k, where (h,k) is the vertex. By substituting the points into the standard quadratic form y = ax^2 + bx + c, we can derive equations: 1. Substituting (2, 3): [ 3 = 4a + 2b + c ] 2. Substituting (8, -1): [ -1 = 64a + 8b + c ] 3. Substituting (11, 8): [ 8 = 121a + 11b + c ] By subtracting the first equation from the other two: - From (8, -1) and (2, 3): [ -4 = 60a + 6b ] [ -2 = 10a + b Rightarrow b = -2 - 10a ] - Substitute b in terms of a into equation from (11, 8) and (2, 3): [ 5 = 117a + 9(-2 - 10a) ] [ 5 = 117a - 18 - 90a ] [ 23 = 27a ] [ a = frac{23}{27} ] [ b = -2 - 10left(frac{23}{27}right) = -2 - frac{230}{27} = -frac{284}{27} ] Substituting a and b back into the first equation to solve for c: [ 3 = 4left(frac{23}{27}right) + 2left(-frac{284}{27}right) + c ] [ 3 = frac{92}{27} - frac{568}{27} + c ] [ c = 3 + frac{476}{27} ] [ c = frac{557}{27} ] Now, knowing a, b, and c, we find the vertex by using the vertex formula h = -frac{b}{2a}: [ h = -frac{-frac{284}{27}}{2 cdot frac{23}{27}} ] [ h = -frac{-frac{284}{27}}{frac{46}{27}} = frac{284}{46} = frac{142}{23} ] Thus, the x-coordinate of the vertex is boxed{frac{142}{23}}.

answer:Let's denote the common ratio of the geometric sequence {a_n} as q. Since a_2a_4=a_5 and a_4=8, we have: a_2^2q^2=a_2q^3 a_2q^2=8 Solving these equations, we find a_2=q=2. Therefore, a_1=1. The sum of the first 4 terms is given by S_4= dfrac{1 times (2^4-1)}{2-1} = 15. Hence, the answers are: q=boxed{2}, S_4=boxed{15}. By setting the common ratio of the geometric sequence {a_n} as q, and from a_2a_4=a_5 and a_4=8, we can derive a_2^2q^2=a_2q^3 and a_2q^2=8. Solving for a_2 and q, and using the sum formula, we can find the solution. This problem tests the understanding of the general term formula and sum formula of a geometric sequence, as well as reasoning and computational skills. It is considered a medium-level question.

answer:Since f(x)= frac {x^{2}+a}{x+1}, then f'(x)= frac {x^{2}+2x-a}{(x+1)^{2}}, Since the slope of the tangent line at x=1 is 1, that is f'(1)=1, then frac {3-a}{4}=1, solving this gives a=-1. Therefore, the correct choice is: boxed{B}. By finding the derivative f'(x) and using f'(1)=1, we can solve for a. This question mainly examines the geometric meaning of the derivative and the basic operations of derivatives, testing the students' computational skills, which is quite fundamental.