Cindy: 4mph
Taylin: 4m/30 min : 8 mph
Judy:5.5m/45min : 7mph
Answer:
See explanation.
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
Functions
- Exponential Property [Rewrite]:

- Exponential Property [Root Rewrite]:
![\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5Bn%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: ![\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bcf%28x%29%5D%20%3D%20c%20%5Ccdot%20f%27%28x%29)
Derivative Property [Addition/Subtraction]: ![\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%20%2B%20g%28x%29%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28x%29%5D%20%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%29%5D)
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: ![\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%20%3Df%27%28g%28x%29%29%20%5Ccdot%20g%27%28x%29)
Step-by-step explanation:
We are given the following and are trying to find the second derivative at <em>x</em> = 2:


We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

When we differentiate this, we must follow the Chain Rule: ![\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%206%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5CBig%5D%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5CBig%5B%20%28x%5E2%20%2B%203y%5E2%29%20%5CBig%5D)
Use the Basic Power Rule:

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%206y%286%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%29%20%5Cbig%5D)
Simplifying it, we have:
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%203%28x%5E2%20%2B%203y%5E2%29%5E%5Cbig%7B%5Cfrac%7B-1%7D%7B2%7D%7D%20%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D)
We can rewrite the 2nd derivative using exponential rules:
![\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202x%20%2B%2036y%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7Bx%5E2%20%2B%203y%5E2%7D%7D)
To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:
![\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%5Cbigg%7C%20%5Climits_%7Bx%20%3D%202%7D%20%3D%20%5Cfrac%7B3%5Cbig%5B%202%282%29%20%2B%2036%282%29%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%20%5Cbig%5D%7D%7B%5Csqrt%7B2%5E2%20%2B%203%282%29%5E2%7D%7D)
When we evaluate this using order of operations, we should obtain our answer:

Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Answer:
See attached image
Step-by-step explanation:
This equation for a parabola is given in vertex form, so it is very simple to extract the coordinates of its vertex, by using the opposite of the number that accompanies the variable "x" in the squared expression (opposite of 2) for the vertex's x-value, and the value of the constant (-6) for the vertex's y-value.
The vertex coordinates are therefore: (-2,-6)
The equation of the axis of symmetry of the parabola is a vertical line passing through the vertex. Since all vertical lines have the shape x = constant in our case, in order to pass through (-2,-6) the vertical line is defined by the equation: x = -2.
See image attached to find the vertex drawn as a red point, and the axis of symmetry as an orange vertical line passing through it.
Answer:
24
Step-by-step explanation: