Is the x2 supposed to be squared?
A.
(0,00
(2.6,7.9)
(4.8,12.4)
(9.7,15.1)
b.
well, the points don't look like they are on the line but they actually are (plot twist)
so
since (0,0) is on th egarph
0=a(0)²+b(0)+c
0=c
so
f(x)=ax²+bx+0 or
f(x)=ax²+bx
find a and b
sub points
(2.6,7.9)
7.9=a(2.6)²+b(2.6)
7.9=6.76a+2.6b
(4.8,12.4)
12.4=a(4.8)²+b(4.8)
12.4=23.04a+4.8b
use those 2 equations
7.9=6.76a+2.6b
12.4=23.04a+4.8b
eliminate b
multiply first equation by -4.8 and 2nd by 2.6 and add them
-37.92=-32.448a-12.48b
32.24=59.904a-12.48b +
-5.68=27.456a+0b
-5.68=27.456a
divide both sides by 27.456
(-5.68/27.456)=a
find b
12.4=23.04a+4.8b
12.4=(-130.8672/27.456)+4.8b
12.4+(130.8672/27.456)=4.8b
(12.4+(130.8672/27.456))/4.8=b
da equation is

c. the roots are found with your calculator
the roots are at x=0 and x=17.287323943662
so very close to the target but not exactly on it
if the target has a radius of 0.287323943662 or more then it will hit the target
1,000,000+3000,000+30,000+4,000+400+50+7
Standard form is ax+by=c where a,b, and c are integers and a is usually positive
so
first find slope
get into form y=mx+b where m is slope
so
slope between points (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
points (-1,5) and (1,2)
slope is (2-5)/(1-(-1))=-3/(1+1)=-3/2
y=-3/2x+b
find b
sub a point
(1,2)
2=-3/2(1)+b
2=-3/2+b
4/2=-3/2+b
7/2=b
y=-3/2x+7/2
add 3/2 both sides
3/2x+y=7/2
times 2 both sides
3x+2y=7 is standard form