We are given the following:
- parabola passes to both (1,0) and (0,1)
 - slope at x = 1 is 4 from the equation of the tangent line 

First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. When x = 0, y = 1. So, c should be equal to 1. The parabola is y = ax^2 + bx + 1 

Now, we can substitute the point (1,0) into the equation,
0 = a(1)^2 + b(1) + 1
0 = a + b + 1
a + b = -1 

The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.
We take the derivative of the equation ,
y = ax^2 + bx + 1
y' = 2ax + b

x = 1, y' = 2
4 = 2a(1) + b
4 = 2a + b 

So, we have two equations and two unknowns, 
2a + b = 4 
a + b = -1

Solving simultaneously,
a = 5 
b = -6

Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .

6 0

3 years ago

Part III: Bailey works part-time at a movie theater. The theater pays employees an hourly wage based on their length of employme

frutty [35]

Given the following table showing the hourly wages of employees of a movie theater based on their length of employment.

Length of employment (months) Hourly wage (dollars/hour)
6 months 7.55
12 months 7.85
18 months 8.15
24 months 8.45

To obtain an algebraic expression that describes how much Bailey’s employer pays part-time workers, we take two points from the table.

Recall that the equation of a linear function satisfying two points
$(x_1,y_1)$ and $(x_2,y_2)$
is given by
$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

Given the two points (6, 7.55) and (12, 7.85), the algebraic expression relating the two points is given by
$\frac{y-7.55}{x-6} = \frac{7.85-7.55}{12-6} = \frac{0.3}{6} \\ \\ 6(y-7.55)=0.3(x-6) \\ \\ 6y-45.3=0.3x-1.8 \\ \\ 0.3x-6y=-43.5$

Therefore, the algebraic expression that describes how much Bailey’s employer pays part-time workers is
$0.3x-6y=-43.5$

5 0

3 years ago

9ab over 27b = simplify 9ab ___ 27b

Darina [25.2K]

Answer:

1A/3 OR 1/3A

Step-by-step explanation:

8 0

3 years ago