Question

WILL GIVE BRAINLIEST!!!!! PLS HELP ASAP!!!! 100 POINTS!!!!!!

Answer 1

Answer:

the linear function gives us the points

(0,3)

(4,0)

(8,-3)

find the equation

hm

find slope

(y2-y1)/(x2-x1)

for (0,3) and (4,0)

slope=(0-3)/(4-0)=-3/4

y=-3/4x+b

given

(0,3)

y=-3/4x+3

so sub -3/4x+3 for y in other equation

(-3/4x+3)^2+x^2=4

expand

9/2x^2-9/2x+9+x^2=4

11/2x^2-9/2x+9=4

times 2 both sides

11x^2-9x+18=8

minus 8 both sides

11x^2-9x+10=0

factor

or use quadratic formula

for

ax^2+bx+c=0

x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}

2a

−b+/−

b

2

−4ac

11x^2-9x+10=0

x=\frac{-(-9)+/- \sqrt{(-9)^2-4(11)(10)} }{2(11)}

2(11)

−(−9)+/−

(−9)

2

−4(11)(10)

x=\frac{9+/- \sqrt{81-440} }{22}

22

9+/−

81−440

x=\frac{9+/- \sqrt{-359} }{22}

22

9+/−

−359

we see we have no real roots

there are no intersection points

#Carry on learning

#Correct me if I'm wrong