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MATH 107 Coursework Assignment: Show Work for All Problems
Coursework Instructions:
SHOW WORK FOR ALL PROBLEMS
- Consider the quadratic equations:
Solve quadratic equation above by
- extracting square roots
- completing the square
- by applying the formula
- Consider the function:
- Will f(x) +4 have real zeros? Justify without any calculations.
- Will f(x) -5 have real zeros? Justify without any calculations.
- Will f(x-2) have real zeros? Justify without any calculations.
- Will f(x+4) have real zeros? Justify without any calculations.
- Consider the function
- Determine the behaviors as and as
- Find all zeros. What is the multiplicity of roots (zeros)?
- When will the function cross the x-axis? When will it cross the y-axis?
- Consider the function
- Find the domain for the function in interval notations
- Find all vertical, horizontal and slant asymptotes
- Sketch the function
- Solve the inequality
Coursework Sample Content Preview:
SHOW WORK FOR ALL PROBLEMS
1 Consider the quadratic equations:
x2-2=0
Solve quadratic equation above by
* extracting square roots
X2-2=0
X2=2
X= = ± √2= {-1.4142, 1.4142}
* completing the square
Reordering the terms:
-2 + X2 = 0
Move all terms containing X to the left, all other terms to the right.
-2 + 2 + X2 = 0 + 2
Combine like terms: -2 + 2 = 0
0 + X2 = 0 + 2
X2 = 0 + 2
Combine like terms: 0 + 2 = 2
X2 = 2
Taking the square root of each side:
X = {-1.4142, 1.4142}
* by applying the formula
Using the formula Quadratic EquationAx2 + Bx + C = 0
A=1, B=0, C=-2
Alternatively, y= x2+ 0x-2
X= −b±√b2−4ac/ 2a
0 ± √02- 4(1) (-2)/ 2(1)
0± √8/ 2
= 0± √4*2/ 2
= 0±2√2/ 2
=0 ±1√2
X= ± √2
2 Consider the function:
fx=x2+4
* Will f(x) +4 have real zeros? Justify without any calculations.
No, the 0’s are solved when f(x) =0 and x is no-negative
* Will f(x) -5 have real zeros? Justify without any calculations.
No, since x is non-negative
* Will f(x-2) have real zeros? Justify without any calculations.
No, this simply shifts the graph rightwards, as f(x-2) is (x-2) ^2+4
* Will f(x+4) have real zeros? Justify without any calculations.
No, f (x+4) simply shifts the graph leftwards, f(x+4) is (x+4) ^2+4
3 Consider the function fx=x5-x3
* Determine the behaviors asx→∞ and as x→-∞
lim x→-∞ (x5-x3)= â&cir...
1 Consider the quadratic equations:
x2-2=0
Solve quadratic equation above by
* extracting square roots
X2-2=0
X2=2
X= = ± √2= {-1.4142, 1.4142}
* completing the square
Reordering the terms:
-2 + X2 = 0
Move all terms containing X to the left, all other terms to the right.
-2 + 2 + X2 = 0 + 2
Combine like terms: -2 + 2 = 0
0 + X2 = 0 + 2
X2 = 0 + 2
Combine like terms: 0 + 2 = 2
X2 = 2
Taking the square root of each side:
X = {-1.4142, 1.4142}
* by applying the formula
Using the formula Quadratic EquationAx2 + Bx + C = 0
A=1, B=0, C=-2
Alternatively, y= x2+ 0x-2
X= −b±√b2−4ac/ 2a
0 ± √02- 4(1) (-2)/ 2(1)
0± √8/ 2
= 0± √4*2/ 2
= 0±2√2/ 2
=0 ±1√2
X= ± √2
2 Consider the function:
fx=x2+4
* Will f(x) +4 have real zeros? Justify without any calculations.
No, the 0’s are solved when f(x) =0 and x is no-negative
* Will f(x) -5 have real zeros? Justify without any calculations.
No, since x is non-negative
* Will f(x-2) have real zeros? Justify without any calculations.
No, this simply shifts the graph rightwards, as f(x-2) is (x-2) ^2+4
* Will f(x+4) have real zeros? Justify without any calculations.
No, f (x+4) simply shifts the graph leftwards, f(x+4) is (x+4) ^2+4
3 Consider the function fx=x5-x3
* Determine the behaviors asx→∞ and as x→-∞
lim x→-∞ (x5-x3)= â&cir...
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