The equations of the asymptotes of the graph of X- 2 are x2 - 4x + 3 the curve y= A x= 1,x = -3,y 1 B x=-1,x = 3,y= 0 /C x=1,x% = 3,y = 0 D x=1,x = 3,y= 1
Q: y = x(x – 1)/3.
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- 7) Determine the x - coordinate of the local maximum of the curve y = (2x-1)^2 (3x+4) in the interval [-1.5, 1].sketch the graph on xyz coordinate y=2 and 3x+2y+z=6sketch the graph of y = x^3 / (x^2 - 4), keep in mind the information provided:1. x and y intercept: (0, 0)2. Vertical asymptotes: x = 2, x = -23. There is no horizontal asymptote.4. Discontinuity at x = -2 and x = 2.5. First derivative: y’ = (x^4 - 12x^2) / (x^2 - 4)^26. Critical points: x = 0, x = 2√3, x = -2√37. Intervals of increase: (-∞, -2√3) U (2√3, ∞)8. Intervals of decrease: (-2√3, -2) U (-2, 0)9. Local maxima: (-3.46, -5.20) and (3.46, 5.20)10. Second derivative: y’’ = (8x^3 + 96x) / (x^2 - 4)^311. Critical numbers: (0, 0), (√12, 3√3), (-√12, -3√3)12. Point of inflection: (0, 0)13. Concave upward: (-2, 0) U (2, ∞)14. Concave downward: (-∞, -2) U (0, 2) A sketch of the graph (by hand) - Labeling of vertical asymptotes on the sketch - Labeling of horizontal asymptotes on the sketch - Labeling of x and y intercepts on the sketch - Labeling of maximum and minimum points on the sketch - Labeling of points of inflection on the sketch A sketch of the graph on Desmos with all key…
- Find the length of the curve with equation x^2 = y^3 between the points (0, 0) and (1, 1).The curve defined by ax^3 + bx^2 - 4x + 12 cuts the x-axis at three points. Two of them are known: 1 and -2. Determine the third.Find a parabola with equation y = ax2 + bx + c that has slope 5 at x = 1, slope −11 at x = −1, and passes through the point (2, 16).
- 11). Find the length (in units) of the curve y = 16 − x2 between x = 0 and x = 2. units PLEASE SHOW STEP BY STEP .Determine the constants a, b, c and d so that the curve defined by y = ax³ + bx² + cx + d has a local maximum at the point (2,4) and a point of inflection at the origin.Alpha particles are deflected along hyperbolic paths when they are directed towards the nuclei of gold atoms. If an alpha particle gets as close as 5 units to the nucleus along a hyperbolic path with asymptote y=3/5x, what is the equation of its path?
- Find all points (if any) of horizontal and vertical tangency to the curve. x = t^2 - t + 9 y = t^3 - 3t (problem attached)If a rectangle has its base on the x-axis and two vertices on the curve y=e^-x^2, show that the rectangle has the largest possible area when the two vertices are at the points of inflection of the curve.I am stuck on the maximum and minimum values of y and corresponding x. Will you please assist on that. It is the last part of the question in the picture.