Final Answer:
a) A polynomial of degree 5 that starts in quadrant II, ends in quadrant IV, has a relative maximum at x = 3, and a relative minimum at x = -3.
b) A linear function with a positive slope that starts in quadrant II, ends in quadrant IV, and has a y-intercept at (0, 0).
c) An exponential function that starts in quadrant II, ends in quadrant IV, and has a horizontal asymptote.
d) A quadratic function that starts in quadrant I, ends in quadrant III, and has its vertex at the origin.
Step-by-step explanation:
a) To sketch the polynomial, start in quadrant II and draw a curve descending to x = -3 (relative minimum), then rising to x = 3 (relative maximum), and finally descending into quadrant IV. Ensure that the total number of turns matches the polynomial's degree, which is 5 in this case.
b) For the linear function, with a positive slope, draw a line ascending from quadrant II to quadrant IV. The y-intercept at (0, 0) means it passes through the origin, ensuring a straight line with a positive slope.
c) The exponential function should start in quadrant II, rise, and asymptotically approach a line in quadrant IV. The horizontal asymptote ensures that as x approaches infinity, the function levels off, never exceeding a certain value.
d) The quadratic function in quadrant I should open upwards, reaching its vertex at the origin, and extend into quadrant III. The symmetry of the parabola ensures that it mirrors on both sides of the y-axis.
In summary, each graph is sketched to meet the specified characteristics, incorporating key features such as relative extrema, asymptotes, slopes, and vertex positions as described in the given instructions.