Free Response Questions
20. A quadratic function has a turning point at (3,8). Selected values for the function are shown in the table
below
(a) Finish filling out the table.
(b) State the zeroes of the function.

Step-by-step explanation:

let the quadratic function be y=ax^2+bx+c

when x=1,y=0

0=a+b+c   ...(1)

when x=3,y=8

8=9a+3b+c   ...(2)

when x=4,y=6

6=16a+4b+c  ...(3)

(2)-(1) gives

8a+2b=8

or 4a+b=4   ...(4)

(3)-(2) gives

7a+b=-2    ...(5)

(4)-(5) gives

-3a=4+2=6

a=-2

from (4)

4(-2)+b=4

b=4+8=12

from (1)

-2+12+c=0

10+c=0

c=-10

so y=-2x²+12x-10

when x=-1,y=-2(-1)²+12(-1)-10=-2-12-10=-24

when x=2,y=-2(2)²+12(2)-10=-8+24-10=6

when x=5,y=-2(5)²+12(5)-10=-50+60-10=0

when x=6,y=-2(6)²+12(6)-10=-72+72-10=-10

b. to find the zeroes

-2x²+12x-10=0

x²-6x+5=0

x²-x-5x+5=0

x(x-5)-1(x-5)=0

(x-5)(x-1)=0

x=5,1

so zeroes are at x=1,5

21.

f(x)=x²+4x-1

=x²+4x+4-4-1

=(x+2)²-5

it is an upward parabola with vertex at(-2,-5)

when x=-6,y=(-6+2)²-5=16-6=11

when x=-5,y=(-5+2)²-5=9-5=4

x=-4, y=(-4+2)²-5=4-5=-1

x=-3,y=(-3+2)²-5=1-5=-4

x=1,y=(1+2)²-5=9-5=4

x=2,y=(2+2)²-5=16-5=11

join these points you get the graph.

f(x) is increasing if f'(x)>0

f'(x)=2x+4

2x+4>0

x>-2

or -2≤x≤2

c. x=0 lies in the interval -2≤x≤2

or in f'(x)>0

also when x=0

f(x)=-1 or in f(x)<0

answer: b.48 for

step-by-step explanation:

50.6

step-by-step explanation: