[SOLVED] Algebra2 Homework 6

Problem 1. Write the following functions in the form f(x) = a(x±h)2±k by completing the square. Describe how x2 is shifted to obtain f(x). Graph f(x), label the vertex, label all axis intersections. An example of what I expect is given below.

(a) f(x)=−2×2 +8x+10 (b) f(x)=3×2 −12x+2

(c) f(x)=−2×2 +3x+2 (d) f(x)=31×2+x+14 (e) f(x)=35×2−3x−41

Example. f(x) = −2×2 − 5x + 1

􏰙

=−2 x +2x +1
5 􏰉5􏰊2 􏰉5􏰊2􏰚

f(x)=−2×2 −5x+1 􏰉2 5􏰊

=−2×2+2x+4−4 +1 􏰙􏰉 5􏰊2 􏰉5􏰊2􏰚

=−2 x+4 − 4 +1 􏰉 5􏰊2 􏰉5􏰊2

=−2 x+4 −(−2) 4 +1 􏰉 5􏰊2 2∗25

=−2 x+4 + 16 +1 􏰉 5􏰊2 25 8

=−2x+4 +8+8 􏰉 5􏰊2 33

=−2x+4 +8
Then f (x) = −2 􏰍x + 5 􏰎2 + 33 is the function x2 shifted left 5 units, stretched vertically by a factor

of 2, reflected about the x-axis, and shifted up 33 units. To find x-intercepts, we set f(x) = 0 and 8

solve for x:

􏰉 5􏰊2 33 −2x+4 +8=0

􏰉 5􏰊2 33 −2x+4 =−8

􏰉5􏰊2 33 x + 4 = −(−2)8

484

􏰉 5􏰊2 33 x + 4 = 16

􏰛􏰉 5􏰊2 􏰜33 x+4 =± 16

√

x + 54 = ±√33

16

√

x+5=± 33 44

√

x=−5± 33 44

√√

33 is positive and −5 − 33 is negative. To find the y-intercept, we set x = 0 and find f(0) = −2(0)2 − 5(0) + 1 = 1. Thus our y-intercept is at y = 1. Noting that our vertex is above the x-axis, on the left of the y-axis, and that the parabola is flipped so that it opens down, it makes sense that one of our x-intercepts is positive and the other is negate. Be sure that all intercepts are labeled and that the vertex is indicated as in the graph below.

Note that −5 +
44 44