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This means that the graph is of form uu thus the vertex is a minimum. x_ ("vertex") will be half way between the x-intecpts x_ ("vertex")= (3+ (-1/2))/2 = 2.5/2=1.25 -> 5/4 Thus y_ ("vertex") = (5/4-3) (5+2) = -7/4xx7=-49/4 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ color (blue) ("Determine the y-intercept") Multiplying out the constants in. So distance (d)= √(x2 −x1)2 +(y2 −y1)2) or d= √(6 − 2)2 +(6 −3)2 = √25 = 5 units [answer] Point a is rotated #pi # clockwise about the origin

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What are the new coordinates of point a and by how much. We know the distance between any two points is the square root of the sum of the x difference squared and the y difference squared The Focus : (3/2,1/2). The eqn. of directrix : x-1=0. graph {x=-y^2+y+1 [-11.25, 11.25, -5.62, 5.62]} :. y^2-y=-x+1 Completing the square we have, y^2-2*1/2*y+ (1/2)^2=-x+1+ (1/2)^2, i.e., (y-1/2)^2=-x+5/4=- (x-5/4)=-4*1/4 (x-5/4). Shifting the origin to the point (5/4,1/2), suppose that, the new co-ords. of (x,y) become (X,Y). Then, x=X+5/4, y=Y+1/2. So, the eqn. in (X,Y) system becomes, Y^2.

In polar form (1 2 + 5 â‹… i) lies on the 1st quardrant

The length of it from the origin is r = √.52 + 52 = √25.25 = 5.025 the argument is θ = tan−1(5.5) or θ = tan−110 = 1.47radian These might be be the actual numbers, but this is not a correct ratio because there is a common factor of #3# (d theta)/(dt) = 37/20 you were on the correct track with your setup on number 8, but when you took the derivative of the y/z term on the right side with respect to time t, you did not use the quotient rule to perform the derivative, which rendered the rest of the problem incorrect The proper setup is sin theta = y/z as you noted

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