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3 years ago

Graph the function f(x) = x + 2 / x − 3 using transformations of the graph of y = 1 / x.

Mathematics

1 answer:

Doss [256]3 years ago

5 0

Answer:

f(1)=1+

−3

Step 1: Simplify both sides of the equation.

f(1)=1+

−3

f=1+2+−3

f=(1+2+−3)(Combine Like Terms)

f=0

Answer:

f=0

Step-by-step explanation:

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SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

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P(x)=R(x)-C(x)
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