Question 431081
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It it is a quadratic in x, so it is a parabola with a vertical axis of symmetry.

The lead coefficient is negative, so it opens downward.

The constant term is 5, so the y-intercept is (0,5)

The opposite of the first degree term is 6, divided by 2 times the lead coefficient, is *[tex \LARGE -\frac{3}{2}].  So the x-coordinate of the vertex is *[tex \LARGE -\frac{3}{2}].  Substitute *[tex \LARGE -\frac{3}{2}] for x and then do the arithmetic to calculate the y-coordinate of the vertex.

By symmetry, there is a function value of 5 a distance of *[tex \LARGE -\frac{3}{2}] horizontally from the vertex, hence (-3,5) is a point on the graph.

Use the discriminant to determine the nature of the zeros.

If there are 1 or 2 real zeros, use the quadratic formula to determine the x-coordinates of the x-intercepts.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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