help big problem?

Question

i am doing my summer assignment for math and i am stuck. i have done 90 other problems and these are the last few that i cant figure out.

where do f and h intersect?
f(x) = 3x-6
h(x)= x^2 -5

(g / f) (-4)
g(x)= e^ (x+4)
f(x) = 3x-6

where do j and k intersect?
j(x)= x- x^3
k(x)= 3x

Answer 1

f(x) = 3x-6
h(x)= x^2 -5

when f & h intersect these two have to be equal
3x-6=x^2-5
x^2-3x+1=0
x={3+-sqrt(9-4)}/2
x=(3+sqrt 5)/2 & (3-sqrt 5)/2
are the values of x at which f(x) & h(x) intersect.

(g/f)(-4)=g(-4)/f(-4)
={e^(-4+4)}/{3*(-4)-6}
=e^0/(-18)
=-1/18...since e^0=1

Again j & k will intersect when
x-x^3=3x
x^3+2x=0
x(x^2+2)=0
x=0 & x^+2=0
when x^2+2=0
x^2=-2
this will give imaginary values of x.
Therefore j & k intersect at x=0 only.

Answer 2

they intersect where f=h
3x-6 = x^2 -5
so
x^2 -3x +1 = 0
using the quadratic formula
x = (3 ± √5) /2
are the two points of intersection

(g/f)(x) = e^ (x+4) /(3x-6)
so
(g/f)(-4)
= e^ (-4+4) /(3(-4) - 6)..........note e^0 = 1
= -1 /18

they intersect where j=k
x- x^3 = 3x
x^3 + 2x = 0
x(x^2 +2) = 0
so when
x=0
and
x^2 +2 = 0
x^2 = -2
x = ± i√2 which are complex numbers, so not a real solution
therefore x=0 is the only point of intersection

the end
.

Answer 3

where do f and h intersect?
f(x) = 3x-6
h(x)= x^2 -5

ans: Where they intersect, f(x) = h(x), and that means

3x - 6 = x^2 - 5 or

x^2 - 3x + 1 = 0 (taking all the terms to one side)

This is a quadratic equation in x of the form ax^2 + bx + c = 0 and has the roots x1= -b + sqrt(b^2 - 4ac) / 2a etc.

Here, a = 1, b = -3 bad c = 1

So, x1 = 3 + sqrt(9 - 4) / 2 = [3 + sqrt5] / 2 and x2 = [3 - sqrt5] / 2


Try the other problems in a similar way.