1.4 Quadratics - Max or Min - Notes [Teacher]

[1.4] QUADRATIC - MAX OR MINSTANDARD FORM (SF)f(x)=ax2+bx+c'c' is the 'y' intercept.To find 'x' intercepts:1. Factor2. Use quadratic Formula:3. Convert to factored form.4. Partial FactoringUse x/h==t to find the axis of symmetry and the 'h' of the vertex (h, k)Axis of Symmetry (AOS): x=hVERTEX FORM (VF)f(x)=a(x-h)2+kVertex = (h, k)Axis of Symmetry (AOS): = hTo find 'y' intercept, you must do f (0)To find 'x' intercepts you can do the following:1. Convert to standard form, then factor or use quadratic formula.2. Convert to factored form.3. UseIf a > 0 then the parabola opens, and it is a MIN.If a <0 then the parabola opens down, and it is a MAX.FACTORED FORM (FF)f(x)=a(x-r)(x-s)r&sarethex-intercepts.Vertex=(h,k)Axis of Symmetry (AOS): x = hEXAMPLE #1: STANDARD FORM TO VERTEX FORM USING PERFECT TRINOMIALSFirst and last terms are perfect squares(a+b)2=a2+2ab+b2Middle term = 2 first termvsecondterm(a-b)2=22-2ab+b2= -1. Convert from Standard Form to Vertex Form to find Vertex.a. x2 + 24x + 144b. x2-18x+8124=122(y)V194=12-12= 9.9fcx) = (x-9) 40V=(-12,0)(44)V=(9,0)(At [1.4] QUADRATIC - MAX OR MINEXAMPLE #2: MAKING PERFECT SQUARES TRINOMIALThe last term of a perfect square trinomial is half of the middle term squared.Factored Form is: (x+))a.x2 + 10x - + 25-(10)21b.x2-24x+144 & (2)2== =2a2(1)2a211)1fes)=(-5)710(-5)+25-2F(12) ( (12) 2-24(12) +144fc-s)=10IV=(-5,0) ! (f(12) =00 IV= (12,0)EXAMPLE #3: STANDARD FORM TO VERTEX FORM USING COMPLETE THE SQUAREf '(x) = =How do I complete the square?1. Put brackets around the first two terms2. Factor out the number in front of the x23. Look at the last term in the brackets, divide it by 2 and then square it.4. Add and subtract that term behind the last term in the brackets.5. Move the negative term outside of the brackets by first multiplying it by the 'a'value.6. Simplify the terms outside of the brackets.-7. Factor the perfect square trinomialConvert the following equation into vertex form (completing the square):f(x) = x2 + 8x + 5(1) = +16,-16= (x248x)+5/5= (x2x8x+16-16)+5-16x11 1(x2+sx+16) - 16)+5+ +4)(x+4) -11VT6 = 4.4=16= (x+4)2-11What is the vertex and axis of symmetry of the parabola? Is the vertex a max or min point?V=(-4,-11)V=(h,k)-- FACES UPAos=-4Hos=X(x)x=h2V=(-4,-11) [1.4] QUADRATIC - MAX OR MINConvert the following equation into vertex form (completing the square):f(x)=2x2-12x+11tout 2= (2x2-12x)+1(2)2-99--9= 2(x2-6x)+112x-9=-18= 2(x2-6x+9) - -18+112(x-3)(x-3) - 7F9=3:3=9(2(x-3)2-7What is the vertex and axis of symmetry of the parabola? Is the vertex a max or min point?IX=3V=(3,-7)- MINAos=3- FACES upV=(3,-7)Convert the following equation into vertex form (completing the square):f(x)=-3x2+9x-13out-3= (-3x2+9x)-13= (3(x"+9x) -13-3x-9= - -= 3(x2+9xt+)-13 27= -3(x-3)(x-1)-25t 2 4What is the vertex and axis of symmetry of the parabola? Is the vertex a max or min point?(x)V=(2)-25)-MAXIX=- FACES(g)Aos= mlDOWN2@2 3 [1.4] QUADRATIC - MAX OR MINEXAMPLE #4: STANDARD FORM - USING PARTIAL FACTOR TO FIND MAX & MINf(x) = ax2 + bx+cHow do I do Partial Factoring?1. Look at the constant term in the expression. Set y = to the constant term.2. Isolate the terms with a variable. Common factor what is left in the expression.3. Solve the remaining equation for X by finding the two places where X is equal tozero. Note: These are not the roots. These are just x-values that have to same y-values.4. Find the axis of symmetry of the parabola by finding the average of the two X-values. You now know the x-value of the vertex.5. Substitute the x-value of the vertex into the original equation and solve for the y-value.a. Use partial factoring to find the vertex. Then state if it is a max or min:x/h = x,+x22fex) = x (x+2)-xth = = -T2X=0x+2=0f(H) X=-1- FACESX=2f(-1) = -7>(x)V=(-lit)y=6b. Use partial factoring to find the vertex. Then state if it is a max or min:f(x) -3x2 +9x-2V=(-1,-7)f(x) = (-3x2+9x) - 2=2f(+) =- 1-3x(x-3)-2((1.5) :-3(15)79(15)-2- 8xx0 X-3-0f(1.s) = 4.75(y) niv=lls, 4.75&x=3(x)-3V=(1.5,4.75)X=0- MAX- FACEDOWM [1.4] QUADRATIC - MAX OR MINEXAMPLE #5: FACTORED FORM - FINDING VERTEXFind the x-intercepts, then determine whether it is a max or min value:f(x) = =3(x-5)(2x+4)fais)= 3(15-5)(2(15)+4)X-5=0 2x+4=0f(is) = - 73.5x=52x=-4-72(g)<=2X=1.5x/h=x,+x2- MIN-FACES= 5+(-2) = 1.5up2EXAMPLE #6 MAX & MIN APPLICATIONSV=(1.5,-73.5)a. The cost of running an assembly line is a function of the number of items per hour. Thecost function is C(x) = 0.28x2 - 1.12x + 2, where C (x) is the cost per hour in thousandsof dollars, and x is the number of items produced in thousands. Determine the economicalproduction level.00 o The Most= = -=2eccnomical2a 2(0.2E) 0.56production levelis22,000/hr.((2)=0.26(2)-1,12(2)+2((2) = 0.88-FACES U- MINV= (2,0.88)(2,0.88)y=2(x)2x1,000=2,000X=20.88x1,000=8805 [1.4] QUADRATIC - MAX OR MINb. Mr. Hayes has 90 apple trees. He earns an annual revenue of $120 per tree. If he plantsmore trees, they have less room to grow, resulting in fewer apples per tree. As a result, theannual revenue, in dollars, is modelled by the function R(x) = (90 + x) (120-x), - where xis the number of additional trees planted. Regardless of the number of trees planted, thecost of maintaining a tree is $8. The cost, in dollars, is modelled by the function C(x)=8(90 + x), where x is the number of additional trees planted. How many trees must Mr.Hayes plan to maximize profit?p(x) = Rcx)- (Cx)P(x) = (90+x) (120-v) - 8(90+x)P(x) = 10,800 - 90x + 120x-x2-720-8xP(x) = -x t ex+19,800x/h = - = - =2a 2(-1): Mr. Me Intos hneeds to plant 11extre trees, so, heP(11)==(11) 722(1)+ 10,080trees to Max profitsshould have 101p(11) > 10,201# OF TREES = 90+11(y)= 101(0,10,080)v=(11,10,201)(x)/16(x=11