ty to the guy who answered!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Here's a few
Task 2
y = x^2 (quadratic)
y = x (linear)
Sub the second equation into the first for y
x = x^2 subtract x from both sides
0 = x^2 - x
x^2 - x = 0 factor
x ( x - 1) =0
Set each factor to 0 and solve for x
x = 0 x - 1 = 0
add 1 to both sides
x = 0 x = 1
So....when x = 0, y = 0
And when x = 1, y = 1
So....the solutions are (0,0) and (1,1)
Here's a graph ;
https://www.desmos.com/calculator/ldpimov7kd
Task 3
y = (x + 3)
______
(x - 2)
The horizontal asymptote occurs at y = 1
The vertcal asymptote occurs at x = 2
The x intercept is at x = -3
The y intercept is at (-3/2)
Here's a graph :
https://www.desmos.com/calculator/4l3egciy58
Task 5
Let the polynomial be x^2 - 4x + 4
Let the linear binomial be ( x - 2)
Part 1
x - 2
x - 2 [ x^2 - 4x + 4 ]
x^2 -2x
_____________
-2x + 4
-2x + 4
_______
0
Part 2
f(a) = f(2) = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0
Part 3
Since we obtained a "0" with polynomial long division......then x = 2 is a root. The Remainder Theorem says that if f(a) = 0, then "a" is a root. This is borne out by the fact that f(2) = 0 where "a" = 2.
Task 6
Let N = 37
We can write this as (x + y) = (49- 12)
So....note that
(49 - 12) ^2 = 49^2 - 2(49*12) + 12^2 = 2401 - 1176 + 144 = 1369 = 372
Task 4
x + a = b
____ ___
ax x
We can "engineer" an extraneous solution, as follows :
Cross-multiply ;
x (x + a) = bax simplify
x^2 + ax = bax
x^2 + ax - bax = 0 let a = 1, let b = 5....so we have
x^2 + 1x - (5)(1)(x) = 0
x^2 + 1x - 5x = 0
x^2 - 4x = 0 factor
x(x - 4) = 0
Set both factors to 0 and solve for x
x = 0 x - 4 =0
x = 4
So.....note that the original problem becomes ;
x + 1 5
____ = __
1x x
Note that if x = 4
4 + 1 5
_____ = ___ which is true
1*4 4
However...if x = 0 ....both denominators are undefined because we cannot divide by 0....thus....x = 0 is an extraneous solution....!!!!!