Answer:
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
