All categories

3 years ago

Consider the absolute value function f (x) = -|x + 2| - 2 . The vertex of the function is .

Mathematics

1 answer:

mezya [45]3 years ago

5 0

The answer would be c
Hope this helps!

You might be interested in

Find the second derivative at the point (1,2), given the function below. y^2-2=2x^3

vagabundo [1.1K]

Solution:

Given:

$y^2-2=2x^3$

Lets First Differentiate the given equation with respect to x

$\frac{d}{dx} ( y^2 - 2 ) = \frac{d}{dx} 2x^3$

$2y \cdot \frac{dy}{dx} - 0 = 6x^2$

$\frac{dy}{dx} = \frac{6x^2}{2y}$

$\frac{dy}{dx} = \frac{3x^2}{y}$ -----------------------(1)

this can be rewritten as

$\frac{dy}{dx} =3x^2y^{-1}$

Now differentiating again with respect to x

$\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{dx}{dy}$

Now substituting (1) we get

$\frac{d^2y}{dx^2} =6x^2y^{-1} + 3x^2 \cdot (-y^{-2}) \cdot \frac{3x^2}{y}$

$\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{3x^2}{y^2}) \cdot \frac{3x^2}{y}$

$\frac{d^2y}{dx^2} = \frac{6x^2}{y} + ( \frac{9x^4}{y^3})$

At(1,2) $\frac{d^2y}{dx^2} = \frac{6(1)^2}{2} + ( \frac{9(1)^4}{(2)^3})$

$\frac{d^2y}{dx^2} = \frac{6}{2} + ( \frac{9}{8})$

$\frac{d^2y}{dx^2} = 3 + 1.125$

$\frac{d^2y}{dx^2} = 4.125$

5 0

3 years ago

You have $100 begging on February 1, 2009, you spend $1 each day. On what day do you spend your last dollar?

Leno4ka [110]

Is it a leap year? haha. But for real your anser will be May 11, 2009

7 0

3 years ago

Read 2 more answers

What is the answer to 8x + 8y x=12 y=-12

Lady bird [3.3K]

Answer:

8x+8y

if the x=12 y=-12

put the value of x and y

8(12)+8(-12)

it turns into 96-96=0

Answer 8x+8y=0

6 0

3 years ago

26 ÷ -6<br><br>as a mixed number and simplified

Fed [463]

Answer:

-4 1/3

Step-by-step explanation:

26 ÷ -6

= 26 / -6 (divide both top and bottom by 2)

= 13 / 3

= -13 / 3

= -4 1/3

7 0

3 years ago

Read 2 more answers

Solve.

jekas [21]

<span>A. {+8) is the answer
</span>

6 0

3 years ago