Hi
f(x) = g(x) if -x²+3x-2 - ( -x+1) = 0
-x² +3x-2 +x-1 = 0
-x² +4x -3 = 0
To solve, tou have to use the general method of resolution of a quadratic fonction.
To determine if it's has a solution in R, let's calculate Δ
Δ = (4)² - 4 * (1) *(-3)
Δ = 16 +12
Δ= 28
as Δ≥ 0 so the function allow two solution within R
so S 1 = ( -4 +√28) / 2 S 2 = (-4 -√28 ) /2
S1 = ( -4 + 2√7) /2 S2 = (-4 - 2√7) /2
S1 = (2 (-2 +√7) /2 S2 2 (-2 -√7) /2
S1 = -2 +√7 S2 = -2 -√7
So the two function are equal twice. one for x = -2 +√7 and second x = -2-√7
The equation of a parallel line will have the same x-term, but a different constant (y-intercept). The required value can be found by putting the given point values in to the equation to see what it needs to be.
... y = 5x + ___
... 8 = 5·4 + ___
... 8 - 20 = -12 = ___
Your equation is ...
... y = 5x -12
Explanation
Step 1
multiplicate by the conjugate
notice that
I hope this helsp you
Interest = p times R times T
here ZLS defines a line segment with end points Z and S.
so it represents line segment ZS
So option B is correct
