# 3.1 Homework.pdf - College Algebra 3.1 Homework 1 Place in...

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Unformatted text preview: College Algebra 3.1 Homework 1. Place in vertex form: f ( x )=x 2−10 x +13. f(x)=x^2 - 10x+25+13 f(x) + 25= x^2 - 10+ 25+13 F(x) + 25= (x+5)^2 +13 -25. -25 f(x) = (x-5)^2 + 13 - 25 f(x)= (x-5)^2-12 2. Place in vertex form: f ( x )=2 x 2+ 12 x−10 . f(x)=2x^2 + 12x-10 f(x)=2(x^2+6x)-10 f(x)=2(x^2 + 6x +9)-10 f(x) + 2.9= 2(x+3)^2 -10 f(x)=2(x+3)^2 -10 -18 f(x)= 2(x+3)^2 - 28 3. Find the vertex and zeros: f ( x )= ( x +6 )2−64 . (x + 6)^2 - 64 = 0 + 64 +64 (x + 6)^2 = 64 x + 6 = √ 64 x+6=8 -6 -6 x= -6 +8 = 2& -14 x- Intercept: (2,0) & (-14,0) y- intercept: f(0) 4. Find the vertex and zeros: f ( x )= ( x −5 )2−8 . (x-5)^2 -8 =0 x-intercept: (7.83,0) & (2.17,0) +8 +8 y-intercept: f(0)= (x-5)^2=8 x-5= +- √ 8 x-5 +- 2√ 2 +5 +5 x= 5 +- 2 √2= 7.83 & 2.17 For the problems, find: 1 a. b. c. d. upward or Downward Graph vertex x-intercepts y-intercept 5. f ( x )= ( x −3 )2−4 a. upward b. f(x)= (x-3)^2 -4 0=((x-3)^2-4 -(x-3)^2=4 (x-3)^2=4 x-3= +-2 x-3=2 x=5 f(0)=(0-3)^2-4 f(0)=(-3)^2-4 f(0)=9-4 f(0)=5 e. f. g. h. sketch the graph axis of symmetry max or min domain & range c. (5,0), (1,0) d. (0,5) f. x=3 g. min Vertex (3,-4) Domain. ( -∞ , ∞ ), {x|x e R} Range. [-4,∞ ) , {y|y≥- -4} 6. f ( x )=2 ( x +1 )2+ 6 A. upward f(0)=2(0+1)^2+6 f(0)=2 1^2 +6 f(0)= 2+6 f(0)=8 y-intercept (8,0) C. x-intercepts none Domain: (-∞ , ∞ ) , {x|x e R} Range [6, ∞ ) , {y|y ≥ 6} vertex:(-1,6) f: x = -1 g. min 7. f ( x )=x 2−4 x−12 A.upward 0= x^2 -4x - 12 x^2 - 4x - 12=0 x^2 + 2x -6x - 12=0 x(x+2)-6(x+2)=0 (x+2)(x-6)=0 x= -2 x=6 C.x-intercepts (6,0), (-2,0) D. y-intercepts: (0, -12) f(0)=0^2 -4(0)-12 f(0)=0-0-12 F(0)= -12 F. x = 2 g. min H. Domain ( - ∞ , ∞ ) {x|x e R} Range: [ -16, ∞ ) , {y|y ≥ -16} 8. f ( x )=3 x 2 +16 x−12 A.upward 0=3x^2+16x-12 3x^2+16x-12 3x^2+18x-2x-12=0 3x(x+6)-2(x+6)=0 (x+6) (3x-2)=0 x=6 x = 2/3 C.(2/3,0) , (-6,0) D. (0, -12) f(0)= 3(0)^2 +16(0)-12 f(0)= 0-12 f(0)=-12 F.x= 2.67 G.min Domain: (-∞,∞), { x|x e R} range:[ - 33.33, ∞) , { y|y ≥ - 33.33 2 9. The monthly profit for a company that makes decorative picture frames depends on the price per frame. The company determines that the profit is approximated by f ( p ) =−8 0 p2 +3440 p−36 , 000 , where p is the price per frame and f(p) is the monthly profit based on that price. (-b/2a, F(-b/2a) a= -80 b= 3440 c= 36000 a. Find the price that generates the maximum profit. (-b/2a, (-3440/ 2x-80) = \$21.50 b. Find the maximum profit. f(p)= 80p^2 + 3440p - 36000 F(21.5)= -80(21.5)^2 + 3400(21.5)-36000 = 980 c. Find the price(s) that would enable the company to break even. f(p)= -80p^2 + 3440p-36000 -80p^2 + 3440p - 36000=0 p^2-43p + 450=0 p^2- 25p - 18p+450p=0 p(p-25)-18(p-25)=0 (p-25)(p-18)=0 10. can be approximated by 2 m ( t ) =−.028 x +2.688 t+35.012 . p-25=0 p-18=0 \$25 and \$18 The gas mileage m(x) for a certain vehicle a. Determine the time at which the population is at a maximum. M(t)= -.028x^2+ 2.688t + 35.012 dm(m)/dx= -0.56x + 2.688 x = 2.688/ 0.56 = 48 mph b. Determine the maximum population. m(t)= -.028(48)^2 + 2.688(48)-35.012 m(t)= 29.5mpg 3 ...
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