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Artist 52 [7]
3 years ago
12

The function f(x) = x2 is transformed to f(x) = −1.4x2. Which statement describes the effect(s) of the transformation on the gra

ph of the original function?
A) The parabola is wider and reflected across the x-axis.
B) The parabola is wider and reflected across the y-axis.
C) The parabola is narrower and reflected across the x-axis.
D) The parabola is narrower and reflected across the y-axis.

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

C) The parabola is narrower and reflected across the x-axis.

Step-by-step explanation:

The original parabola has equation:

f(x) =  {x}^{2}

The transformed parabola has equation

f(x) =  - 1.4 {x}^{2}

How wide the graph is can be determined by the absolute value of the coefficient.

The smaller the absolute value of the coefficient, the wider the graph.

Since

|1|  \:  <  \:  | - 1.4|

The original graph is wider than the transformed graph.

Also the negative factor tells us there is a reflection in the x-axis.

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A ball is thrown up in the air, and it's height (in feet) as a function of time (in seconds) can be written as h(t) = -16t2 + 32
olga2289 [7]

Answer:

1) Using properties of the quadratic equation:

Here the height equation is:

h(t) = -16t^2 + 32t + 6

We can see that the leading coefficient is negative, this means that the arms of the graph will open downwards.

Then, the vertex of the quadratic equation will be the maximum.

Remember that for a general quadratic equation:

a*x^2 + b*x + c = y

the x-value of the vertex is:

x = -b/2a

Then in our case, the vertex is at:

t = -32/(2*-16) = 1

The maximum height will be the height equation evaluated in this time:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

2) Second method, using physics.

We know that an object reaches its maximum height when the velocity is equal to zero (the velocity equal to zero means that, at this point, the object stops going upwards).

If the height equation is:

h(t) = -16*t^2 + 32*t + 6

the velocity equation is the first derivation of h(t)

Remember that for a function f(x) = a*x^n

we have that:

df(x)/dx = n*a*x^(n-1)

Then:

v(t) = dh(t)/dt = 2*(-16)*t + 32 + 0

v(t) = -32*t + 32

Now we need to find the value of t such that the velocity is equal to zero:

v(t) = 0 = -32*t + 32

       32*t = 32

            t = 32/32 = 1

So the maximum height is at t = 1

(same as before)

Now we just need to evaluate the height equation in t = 1:

h(1) = -16*1^2 + 32*1 + 6 = 22

The maximum height is 22ft

4 0
3 years ago
A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How far does the ju
weeeeeb [17]

Answer:

x = 9.5 m

Step-by-step explanation:

Given:

This problem is based on 2D motion kinematics equations:

V_0 = 12 m/s

\theta = 20^0

g=9.8 m/s^2

At maximum height:

V_y = 0

Components of initial velocity:

V_0x = V_0 cos (\theta) =(12 )(cos (20^0)) = 11.3 m/s

V_0x = V_0 sin (\theta) = (12)(sin (20^0)) = 4.10 m/s

To find time to reach maximum height:

V_y = V_0y -gt

Plugging in the known values:

0 = 4.10 - 9.8t

t = \frac{-4.10}{-9.8}

t = 0.418 s

Total time = 2t = 2(0.418) = 0.836 s

to find horizontal distance covered:

x = (V_0x)(t)\\x = (11.3)(0.836)\\x = 9.5 m

8 0
3 years ago
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