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3 years ago

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Consider the graphs of j(x)== sin (6x- pi 2 )+11 and k(x) = 7sin(6x - pi/2) - 11 Which feature of the graph of k(x) is different

from the graph of j(x) ? the amplitude the frequency the phase shift the vertical shift

1 answer:

Tcecarenko [31]3 years ago

4 0

Answer:

The answer is D

You typed the first equation wrong but i just did this on edgenuity (with the right equations) and got it right

Step-by-step explanation:

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Read 2 more answers

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes

ArbitrLikvidat [17]

Complete Question

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for $x = 3$ and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ?

Answer:

The generally expression for the slope of $y = 2x^2 + 4x$ is $y' = 4x +4$

The graph is shown on the first uploaded image

The generally expression for the slope of $y=2x^2+4$ is $y' = 4x$

Step-by-step explanation:

From the question we are told that

The equation of the curve is $y = 2x^2 + 4x$

First we differentiate the equation

So

$y' = 4x +4$

Therefore the generally expression for the slope tangent to the curve $y=2x^2+4x$ is $y' = 4x +4$

The next step is to substitute for x = 3 and x = 0.5

So for $x_1 = 3$

$y' = 4(3) +4$

$y' =m_1= 16$

And for $x_2 = 0.5$

$y' = 4(0.5) +4$

$y' =m_2= 6$

Here m_1 and m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So at $x_1 = 3$

$y_1 = 2(3)^2 + 4(3)$

$y_1 = 21$

So the coordinate for the first tangent line is

$(x_1 , y_1 ) = (3 , 21)$

At $x_2 = 0.5$

$y_2 = 2(0.5)^2 + 4(0.5)$

=> $y_2 = 2.5$

So the coordinate for the second tangent line is

$(x_2 , y_2 ) = (0.5 , 2.5)$

Next we obtain the equation for the tangent lines

So generally the slope is mathematically represented as

$m = \frac{y - y_1 }{x-x_1}$

For $(x_1 , y_1 ) = (-3 , 21)$ and $y' =m_1= 16$

$16 = \frac{y -21 }{x-3)}$

=> $y = 16x - 27$

For $(x_2 , y_2 ) = (0.5 , 2.5)$ and $y' =m_2= 6$

$6 = \frac{y -2.5 }{x-0.5}$

$y = 6x -0.5$

Generally the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) is mathematically evaluated by differentiating y=2x^2+4 as follows

$y' = 4x$

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