To help you find the range of a quadratic function, look at the of the parabola.

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

Heather is writing a quadratic function that represents a parabola that touches but does not cross the x-axis at x = -6. which function could heather be writing? fx) = x2 + 36x + 12 = x2 - 36x - 12 f(x) = -x + 12x + 36 f(x) = -x? - 12x - 36

Segment wx is shown explain how you would construct a perpendicular bisector of wx using a compass and a straightedge