"Unit 2 ProjectComplete parts a­c for each quadratic function.a. Find the y­ intercept, the equation of the axis of symmetry, and the x­ coordinates ofthe vertex.b. Make a table of values that includes the vertex.c. Use the information to graph the function.2 1. f(x) = ­3x + 8xSolutions:a. y-intercepts is (0,0)3x+8=03x=-8 subtract 8 from both sides of an equations)3x=-8/3=2.667 therefore: y intercept is (0,0) x=-8/3=-2.667b. Table:2f(x)=-3x +8xX Y 0 0 -8/3 02.6670 c. Graph: y-intercept( -8/3,0) (0,0)(2.667,0). ..-3-2 -1-112 3 -2 -32 2. f(x) = 2x + 7x + 1Solutions: 2a. x-axis intercept point of 2x +7x+1; (-7 + v4 1 ,0 ),(-7- v4 1 ,0 )4 4 y-intercept=(0,1) b. Table of values that includes vertex: x y -7+v41 0 401-7-4v410 4 c. graph: 4 2-42(0,1) . (-7+4v41) (-7-4v41 4 2 4 4-2 -4Determine whether each function has a maximum or minimum value. State themaximum or minimum value of each function.2 3. f(x) = x + 6x + 9 Answer: Minimum is (-3,0)2 4. f(x) = ­x + 4x Answer: Maximum is (2,4)5. Write a quadratic equation with roots ­3 and 4 in standard form. Solution: If x=-3, then x+3=0If x=4, then x-4=02 So..(x+3)(x-4) will foil as x - 4x +3x-12=0Solve each quadratic equation using the method of your choice. Find exactsolutions.2 6. ­1.6x ­ 3.2x + 18 = 0Solutions: 2x=-b+vb -4ac2x= -3.2+v(3.2) -4(1.6)(18)2(1.6) x=3.2+v-104.96 3.22 7. 10x + 3x = 1 Solutions:Step 1; Subtract 1 from both sides210x +3x-1=1-1210x +3x-1=0Step 2; Use the quadratic formula with a=10, b=3, c= -12 x=-b+vb -4ac2a2x=-(3)+v(3) -4(10)(-1) 2(10) x=-3+v4920 x=1/5 or x=-1/2 2 8. x + 8x ­ 48 = 0Solutions:2 Use the quadratic formula x=-b+vb -4ac2a2x=-(8)+v(8) -4(1)(-48)2(1) x=-8+v256 2(1) x=4 or x=-12Simplify the expression.9. (5 ­ 2i) ­ (8 ­ 11i) Solutions: Just group the real part and the imaginary part; (5-8)+(-2+11)i (5-2i)-(8-11i)=-3+9i2 10. (14 ­ 5i)Solutions:Apply distributing rule2 . . 2 14 -2 14 5i+(5i) : 171-140i Write each equation in vertex form. Then identify the vertex, axis of symmetry,and direction of opening.2 11. y = x + 10x + 27Solution:21(x +(10)x)+(27)21(x +(10)x+25-25)+(27)21((x+(5) -25)+2721(x+(5)) +2)Therefore: The vertex is (-5,2) The axis is x=-5 Direction of opening; The parabola opens up. 2 12. y = ­9x + 54x ­ 8Solutions:. 2-9 (x +(-6)x+9-9)+(-8). 2-9 ((x+(-3)) -9)+(-8). 2-9 (x+(-3)) -(-81)+(8). 2-9 (x+(-3)) +(73) Therefore:The vertex is (3,73)The axis is x=3Direction opening; The parabola opens down.Graph each inequality.13. (x ­ 5)(x + 7) < 0? -7<x<5-7 5 xAnswer: {x-5=0 {x+7=0 {(x-5)+5=5 {(x+7)+(-7)=-7 Results: x € (-7,5)2 14. ­5x + x + 2 < 0Answer: ? x<-2/11 -0.18182-5-4-3-2 -1 0 1 2 34 .5 2x+x+2< 0 10x+x+2< 0 result: x<-2/11Find the exact solution to the system of equations. Check your answeralgebraically.2 15. y = x ­ 6x + 1 y+ 2x = 6 Solutions: Set these equations equal and find the value of x2y=x -6x+1y+2x=32 X -6x= 2x-6subtract (2x-6) from both sides2 x -6x-(2x-6)=2x-6-(2x-6) refine2 x -8x+6=02 Now solve with the quadratic formula: a=-b+vb -4ac 2a2 . . x=-(-8)+v(-8) -4 1 6: 4+v10. 2 1 therefore x=4+v10, or x=-4v10 algebraically check:2x -6x=2x-6 @ 4+v102 ( 4+v10) -6(4+v10)=2(4+v10)-6 8.324555=8.324555 TRUE 2 16. 2x ­ 4x = y + 1 x+ y = 1Solutions;Set these equations equal and find the value of x22x -4x=y+ 1x+y=122x -4x=1x-1solve with the quadratic formula: 2 . 2x -4x=1 x-1: 5+v174(decimal: x=2.28078, or x=0.21922)Steps:2 . 2x -4x=1 x1: x-12 2x -4x=x-12 2x -4x(x-1)=x-1(x-1) Refine:2 2x -5x+1=02Solve for quadratic formula: x=-b+vb -4ac 2a2 . .x=-(-5)+v(-5) -4 2 12(2)Therefore x=2.28078, or x=0.21922Algebraically check:2 2x -5x+1=0@ x=2.280782 2(2.28078) -5(2.28078)+1=00.000015 ?0TRUE 2 2x -5x+1=0 @ x=0.219222 2(0.21922) -5(0.21922)+1=0 0.000015 ?0TRUE "