Equations don't really have x intercepts, functions or curves do. They're also known as the zeros of the function. When we set a function equal to zero we get an equation to solve, and the zeros of the function become the solutions or roots of the equation.
If a quadratic equation only has one root, that's a repeated root corresponding to a discriminant of zero.
In this example our equation is something like 
, or expanded
The discriminant 
here is
Answer:
Step-by-step explanation:
Step One: Convert
49/16x^2-2=-0.05x+4
Step Two: Multiply Both Sides by 80
245x^2-160=-4x+320
Step Three: Move everything to the left
245x^2+4x-480=0
Final Step: Quadratic Formula
(-1.5,3) Y=3 which is up 3 and a horizontal line and 2x+y=0 can be written as y=-2x so in order to get -1.5 you have to make y=3 cause y=3 and when you do that it should look like 3/-2=-2/-2x so x=-1.5
It has a constant daily growth rate of 1.5
Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
