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amid [387]
2 years ago
13

which statement correctly descibes what happens to the shape and location of the graph of y=2x^2 if it is regraphed as y=-1/2x^2

Mathematics
1 answer:
lbvjy [14]2 years ago
4 0
I dont have options but i can sayEach of these questions is asking you to figure out how g(x) would be graphed based on some other function, f(x), except we're gonna make some changes to f(x) first. For example, let's look at part (a).  You're told that g(x) = 2f(x)+3.  1. Whatever f(x) is, 2f(x) would take all of the y values and double them.2. Whatever that turns out to be, "+3" would add 3 to every y value. So, to translate (1, -2), first we double the y value...so that'd be -2 * 2, or -4.  Then we add 3, which gives 1.  So, the point (1, -2) becomes (1, 1).  [EDIT:  Um, no, -4 + 1 isn't 1.  It's -1.  My bad!  MBW] Part (b) is a little bit trickier: 1.  Whatever f(x) is, by adding 1 to x, it actually shifts the graph left by 1, even though it sorta feels like you should be shifting it 1 to the right.  Let's not get too far into the details here, so for the moment, f(x+1) means "shift the graph of f to the left by 1."  In other words, subtract 1 from x to get the new point. 2.  Whatever f(x+1) is, then the "-3" would subtract 3 to every y value. So, if (1, -2) was on the original graph, then f(x+1) would be shifted to the left by 1...so that's (0, -2).  And then, we subtract 3 from y, so that'd be (0, -5).  Basically, anything inside the parentheses, like f(x+1), messes with the x coordinate of the point, and anything outside the parentheses, like -f(x) or f(x) + 3, messes with the y coordinate. For the last two, I'll give you a few hints and see if you can take it from there. For part (c):  f(2x), even though you think it might double the value of x, actually divides the value of x by 2.For part (d):  the trick here is to ignore the "-x" until the last step.  Deal with the f(x-1) first (which shifts the x coordinate...which way?), then the negative outside of f(x) (which flips the sign of the y coordinate), then the +3 outside of f(x)...and then, at the end, the "-x" would flip the sign of the x coordinate.  I hope this helps point you in the right direction!

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2 years ago
Prove the following
fomenos

Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
  • sin (270°-θ)  =  -cos θ
  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
  • cot (270°+θ)  =  -tan θ
7 0
2 years ago
Read 2 more answers
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