how do you find the algebrically, the point(s) of intersection of:?

Question

(a) f(x) = x+4 and f(x) = 3x-8
(b) f(x) = x2+2x-8 and f(x) = 3x-2

Answer 1

a) We want to find the values of x which give equal values of f(x) for both functions. So set x+4 = 3x-8 and solve for x:

x+4 = 3x-8
4 = 2x-8
12 = 2x
x = 6

Plugging this back into f(x) for either equation, we get f(6)=10. So the point of intersection is (6,10).

b) Same strategy:
x² + 2x - 8 = 3x-2
x² - x - 8 = -2
x² - x - 6 = 0
(x - 3)(x + 2) = 0

In this case there are two points of intersection, the x values of which are x = 3 and x=-2. These give 7 and -8 for the f(x) values, respectively. So the points of intersection are (3,7) and (-2,-8).

Answer 2

a.
f(x) = x + 4
f(x) = 3x - 2
x + 4 = 3x -2
x- 3x = -2 - 4
-2x= -6
-x = -3
x = 3
b.
f(x)= x2 + 2x-8 & f(x) = 3x -2
x2 + 2x -3x = -2+8
x2 -x = 6
x2 -x -6
(x -3) ( x +2)
x= 3 or x = -2
f(3)= 3+4=7
f(-2)= -8

Answer 3

They intersect when they are equal...

so for a)
x+4 = 3x - 8
x = 3x - 12
-2x = -12
x = 6

Then put 6 back into one function to find f(x) (otherwise known as "y").
x + 4 = 6 + 4 = 10

the point is (6, 10)
-----
b)
x^2 + 2x - 8 = 3x - 2
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3, -2

f(3) = 3(3) - 2 = 9 - 2 = 7
f(-2) = 3(-2) - 2 = -6 - 2 = -8

So: (3, 7) and (-2, -8) are the points.

Answer 4

(a)
x + 4 = 3x - 8
2x = 12
x = 6
f(x) = 10
p = (6, 10)

(b)
x^2 + 2x - 8 = 3x - 2
x^2 -x - 6 = 0
(x + 2)(x - 3) = 0
x = -2, 3
f(x) = -8, 7

Answer 5

Solve the system of equations. You can do this with an augmented matrix or substitution.