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3 years ago

What is the average rate of change for the function f(x)=x^2+4 for the domain of -2 to 4

Mathematics

1 answer:

Neko [114]3 years ago

5 0

Answer:

Step-by-step explanation:

For a quadratic function the average rate of change on an interval is the rate of change at the midpoint of the interval. The rate of change of a function is given by its derivative.

The derivative of f(x) = x^2 is f'(x) = 2x. The midpoint of the interval is (4+(-2))/2 = 1. Then the average rate of change is ...

f'(1) = 2(1) = 2

The average rate of change of f(x) on [-2, 4] is 2.

Alternate solution

The average rate of change is the slope of the line between the end points of the interval:

m = (y2 -y1)/(x2 -x1)

m = (f(4) -f(-2))/(4 -(-2)) = (20 -8)/(6) = 2

The average rate of change on [-2, 4] is 2.

_____

The attached graph shows the points on the curve and a line with slope 2 between them. It also shows the various slope calculations.

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alexgriva [62]

Answer:

-1 & -2

Step-by-step explanation:

x²+3x+2

x²+2x+x+2

(x²+2x)+(x+2)

x(x+2)+1(x+2)

(x+1)(x+2)

6 0

3 years ago

Read 2 more answers

Mark uses an app that shows him how many kilometers he has run to prepare for a marathon. The app said he ran 9.654 kilometers.

Viefleur [7K]

Answer:

5.9987174899

Step-by-step explanation:

9.654km / 1.609344

= 5.9987174899mi

4 0

3 years ago

Write a quadratic function to model the vertical motion for each situation, given h(t) equals negative 16t squared plus v0t+h0

Mashcka [7]

Answer:

hmax = 194 ft

The maximum height is 194 ft

Step-by-step explanation:

According to the given equation for the model of the vertical motion. The height at any point in time can be written as;

h(t) = -16t^2 + v0t + h0 .......1

Where;

h(t) = height at time t

t = time

v0 = initial velocity = 96 ft/s

h0 = initial height = 50 ft

To determine the maximum height we need to differentiate the equation 1 to find the time at which it reaches maximum height;

At the highest point/height h' = dh/dt = 0

h'(t) = -32t +v0 = 0

-32t + v0 = 0

t = v0/32

t = 96/32

t = 3 s

At t=3 it is at maximum height.

The maximum height can be derived from equation 1;

Substituting the values of t,v0,h0 into equation 1;

h(t) = -16t^2 + v0t + h0 .......1

hmax = -16(3)^2 + 96(3) + 50 = 194 ft

hmax = 194 ft

The maximum height is 194 ft

5 0

3 years ago

Evaluate c-2c−2c, minus, 2 when c=7c=7c, equals, 7.

____ [38]

Answer:

the correct answer would be 5

Step-by-step explanation:

when c=7, then we put that value in for c in the equation c-2

c-2

(7)-2

3 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: