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0

Because the vertex of a parabola is the point $(\bar{x},\bar{y})$ such that $\bar{y}$ is maximum (if $a < 0$) or minimum (if $a > 0$), and these situations both happen for $\bar{x} = h$, which ultimately leads to $\bar{y} = k$. You can also think of the symmetry of the parabola in the $y$-axis. Meaning, find the roots of $a(x-h)^2+k$ and take their ...


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A straight parabola (i.e., one with its directrix parallel to the $\;y$- axis) has its maximal or minimal point at its vertex (depending on its leading coefficient's sign), and $$f(x)=y=(x-h)^2+k\implies f'(x)=2(x-h)=0\iff \color{red}{x=h}$$ and then $$f(h)=k$$ so the vertex is indeed at $\;(h,k)\;$ , and it is a minimum point iff ...


3

As pointed out in the comments, the graph is indeed an ellipsoid. In the event you don't have a 3D graphing program handy, here's one way to think about it: we have the equation $$x^2+2y^2+3z^2=12.$$ We'll set the coordinates $x,y,z=0$ (one at a time) and this will tell us what the graph looks like in the $yz$, $xz$, and $xy$-planes respectively. For ...


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It is an ellipsoid. Source: http://www.wolframalpha.com/input/?i=x^2%2B2y^2%2B3z^2%3D12


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Yes, it means that you will only consider values of $t$ ranging from $-2$ to $3$, although the expressions make sense for all $t$. Notice that $-2 \leq t \leq 3 \implies -4\leq 2t \leq 6$, that is, $-4 \leq x \leq 6$. Eliminating $t$ from these expressions, we obtain $$y = \frac{x}{2}+5,$$ a line segment in the plane, joining the points $(-4,3)$ and $(6,8)$. ...


2

Your graph is correct. The vertical segments you see in the second graph may be a result of a graphing program that is set in continuous mode so that it connects the dots even when they should not be connected, as in a step function or at a vertical asymptote. A function $f$ associates a unique value $f(x)$ to each $x$ in its domain. When we graph a ...


1

In a large amount of cases, converting the function into polar coordinates works very well. For example, for $f(x_1,x_2) = \frac{x_1x_2}{\sqrt{x_1^2+x_2^2}}$ converts into $$f(r,\phi) = \frac{r^2\cos\phi\sin\phi}{r}$$ and the limit as $r\to 0$ is easy to calculate.


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your $u$ has a horizontal asymptote. so you you know the derivative should have a horizontal asymptote too. you derivative will have a positive local max and then decrease and approach zero. your derivative should look a little bit like the reflection on the $x$-axis of $u.$


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Suppose that $f(1) = 1, f(10) = 2, f(100) = 3.$ Let's suppose further that you measure position on your paper in centimeters, with the origin being at the origin of your graph. If you plot $\log(x)$ vs $f(x)$, you'll plot points at $(0cm, 1cm), (1cm, 2cm),$ and $(2cm, 3cm)$. If, on the other hand, you use the log paper's log-scale on the x-axis, let's ...


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This example isn't too great if you're looking for a smooth, simple function (since it's an oscillating, discrete one), but I believe it conveys the 'physical idea' of vertical asymptotic behavior nicely, though it's doesn't actually involve one at all. Imagine a lamp that is on for 1s, then off for 0.5s, then on for 0.25s, then off for 0.125s, and so on; ...


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So with the points (1,3000), (2,9000), (3,18000), and (4, 30000), I was able to use Mathematica to find a polynomial to fit the trend. It came to $y = 1500x+1500x^2$. You could also use Microsoft Excel if you do not have Mathematica. If you wanted to do this by hand, then you could use the method of constructing what is known as a Lagrange polynomial. ...


9

Physics has lots of examples, but it's already the closest field to math. (Plus, in order to observe asymptotic gravity, you'd need a black hole...) You could use Walmart. If shoppers arrive nondeterministically at rate $\lambda$ and are served at nondeterministically at rate $\mu$, the average wait time is $$\frac{1}{\mu − \lambda} − \frac{1}{\mu}$$ ...


2

Have you tried to describe, how high is the aiming point at the wall in front of you when you rise a rifle at a given angle? (answer: $\mathrm{height} = \mathrm{distance} \times \tan(\text{angle})$ with a vertical ;) asymptote at $\tfrac \pi 2$) Another 'tan' disguise: how far away from the Earth you need to be to see half of it? Similar: $\sec x =\tfrac ...


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Newton's inverse square law of gravity! Can't much closer to real-life, everyday experiences than gravity! The law reads: $F = G\frac {m_1m_2} {d^2}$ Where $F$ is the gravitational force between two bodies, $G$ is the universal gravitational constant, $m_1$ and $m_2$ are the masses of the two bodies, and $d$ is the distance between the two bodies. ...


2

I suspect there is no natural sigmoid function that will "preserve" in some sense your original function $f$. In all generality, you want to "squeeze" the real line ($\mathbb{R}$) into an (open) interval of the form $(x_1,x_2)$. Let's try to squeeze both the $x$-axis and the $y$-axis so that we may fit one graph into a rectangle. Let this rectangle be, for ...


1

There is an example from physics. The tension in a rope hung between two trees. http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W03D1-1.pdf In order for the rope to be absolutely flat, the tension at the ends must be infinite.


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Slamming the brakes or even crashing in a car? Gently -> feel it a tiny bit Very hard -> get jerked around Crash head-on -> die I'm not thinking too hard about the mathematical model but it's exactly the right concept in that as time goes to zero, physical damage explodes "infinitely."


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Throw a stone obliquely. Due to air friction, the trajectory follows a vertical asymptote. http://www.mathcurve.com/courbes2d/paraboleamortie/paraboleamortie.shtml


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From distance equals rate times time, you get $r=\frac{d}{t}$ For a fixed distance, the less time you take to cover that distance, the faster you go, with a vertical asymptote at $t=0$.


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You will need more than two pairs to do a least square fit. Transform the equation to $$\ln{y}=a-b\ln{x}$$ Then you can use usual least square method to find the parameter $a,-b$, using the pairs $(\ln{x_i},\ln{y_i})$.


11

One example would be the gravitational potential energy of a point in relation to a pointwise mass in space. The closer you are to the point, the faster you go. http://en.wikipedia.org/wiki/Potential_energy#Potential_energy_for_gravitational_forces_between_two_bodies If you want simpler examples, take any basic equation that implies a linear connection ...


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question A: $f'(-2) > 0$ Yes, it is correct. It looks like there is an instantaneous slope of around $+1$. question B: $f(0) = 0$ and $f'(0) = 0$ and $f''(0) \neq 0 $ No, it is not correct because the statement $f''(0) \neq 0$ is false. (The statements $f(0)=0$ and $f'(0)=0$ are true, by the way.) From the graph, you will see that concavity changes. ...


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(Thinking of $a>0$, as it seems that you do) You seem to be OK (correct me if I'm wrong) that $f(x-a)$ shifts the graph of $x\mapsto f(x)$ to the right with $a$ steps. OK? Thus, to shift the function $x\mapsto f(-x)$ to the right $a$ steps, we want $f(-(x-a))=f(a-x)$.


1

Define the graph of a function $f:D\rightarrow\mathbb R$ to be the set of points $$ G(f)=\{(x,f(x))\ \mid\ x\in D\} $$ Then the function $g(x)=f(x-a)$ is defined whenever $x-a=y\in D$ so that $x\in E=\{y+a\ \mid y\in D\}$ and the graph is $$ \begin{align} G(g)&=\{(x,g(x))\ \mid\ x\in E\}\\ &=\{(x,f(x-a))\ \mid\ x\in E\}\\ &=\{(y+a,f((y+a)-a))\ ...


0

The graph of $x\mapsto f(x+a)$ is just the composition of $f$ with the translation $x\mapsto x+a$. Therefore, if $D$ is the domain of $f$, the graph of $x\mapsto f(x+a)$ is the same graph than $f$ but on the domain $D+a$.


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Meanwhile, I figured it out. Actually, the solution is quite easy and natural: common := coords = cylindrical: plots:-display([ plot3d([1, phi, z], phi = 0..Pi/3, z = 0..2, common), plot3d([2, phi, z], phi = 0..Pi/3, z = 0..2, common), plot3d([r, phi, 0], r = 1..2, phi = 0..Pi/3, common), plot3d([r, phi, 2], r = 1..2, phi = ...


0

For universe $U$ and unit metric $m$ defined over $\mathbb{R}^1$, the equation $U = \frac{x^2}{m}$ holds for all values such that $x=\sqrt {42}$.


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Recall the definition of the slope of a line. The slope of a line that passes through the points $(x_1, y_1)$ and $(x_2, y_2)$ is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Since we cannot divide by zero, the slope is only defined if $x_1 \neq x_2$, that is, the slope is only defined if a line is not vertical. Suppose that a line non-vertical passes through ...


1

Pick a point $t$ on horizontal axis, and look at it's value $f(t)$ . As you add a non zero number to $t$ to the horizontal axis, say $1$, if you look at the value in that point $f(t+1)$, you are only adding a multiple of that number to vertical axis - $f(t+1)=c*(t+1)=c*t + c*1$. If the constant is $0$, you are not adding anything as you go along the ...


1

Let $A(x_1, y_1)$ and $B(x_2, y_2)$. The line from (4, 17) to (11, 3) has a slope of $\frac{17-3}{4-11}=\frac{14}{-7}=-2$. Thus the line AB has a slope $s$ satisfying the following equation: $-2*s=-1$ (since the lines are perpendicular). This leads to $s=\frac{1}{2}$. The points A and B can then be found by starting at (11, 3) and just using our slope of ...


1

The slope between the given points is $$ m = \frac{3 - 17}{11 - 4} = -2 $$ The slope of the perpendicular line is $$ m' = -\frac{1}{m} = \frac 12 $$ Define $$ \Delta x = \cos(\arctan(m')) = \frac{1}{\sqrt{(m')^2 + 1}} = \frac{2}{\sqrt{5}}\\ \Delta y = \sin(\arctan(m')) = \frac{m'}{\sqrt{(m')^2 + 1}} = \frac{1}{\sqrt{5}} $$ The coordinates of the points you ...


0

HINT: For $|2i-z|=|z+1+3i|$: Distance from $(0,2)$ = Distance from $(-1,-3)$ in $\mathbb{C}$. For $Re[(3-4i)z]>0$: If you visualize the geometric notion of multiplication of $\mathbb{C}$ numbers and try to keep the product have a positive $X$, you will realize that $z$ lies in the half plane just missing the origin and farthest from the point $(-3,4)$. ...


1

$|2i-z|$ represents the distance of $z$ from $2i$ and $|z+1+3i|$ represents the distance of $z$ from $-1-3i$. If the distance is the same then draw a line connecting these $2$ points. And then in the middle of this line draw another line perpendicular to it. This line represents your set. For $z = x+yi$, $(3-4i)z = 3x+3yi-4xi + 4y$. So $Re((3-4i)z) > 0 ...


0

Well, in order to determine the shape of the graph, you need to understand that the [(x^2/a) -(y^2)/b = z/c] will result in a saddle shape or a hyperbolic paraboloid. Basically, the crest, or bottom, of the level curve will be at (0,0,0) and there will a parabolic curve on the z-x plane with a decline on the 3D space:


0

You need a software that solves ODEs numerically. Mathematica code is given that produces same graphs: {a, b, c, d} = {.1, .01, .05, .001}; NDSolve[{X'[t] == a X[t] - b X[t] Y[t], X[0] == 50, Y'[t] == d X[t] Y[t] - c Y[t], Y[0] == 15}, {X, Y}, {t, 0, 500}]; {x[u_], y[u_]} = {X[u], Y[u]} /. First[%]; Plot[{x[t], y[t]}, {t, 0, 500}, AspectRatio -> ...


1

When $\Delta_t=0$, and $N_t > 1$, $G=1$ as well.


0

The curve obtained by combining the graphs of $y = \sqrt{x}$ and $y = -\sqrt{x}$ is $x = y^2$. It is not a function of $x$ since there are two values of $y$ for each $x > 0$. However, you can use the parametric equations \begin{align*} x(t) & = t^2\\ y(t) & = t \end{align*} to write both $x$ and $y$ as functions of a third variable $t$. Observe ...


6

What you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a "sideways parabola," since the "standard" parabola that people learn opens towards the positive $y$ axis. The bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on ...


0

If you want the square function $$ y = x^2 $$ turned 90 degrees to the right, you get the inverse $$ x = y^2 $$ or written in another way (like you already mentioned) $$ y = \pm \sqrt{x}. $$ I'm not actually sure what you mean by writing it as a single function as it has two branches (the positive and the negative one).


4

The graph you describe will have equation $x=y^2$. You cannot write it in the form "$y=\langle\hbox{function of $x$}\rangle$" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.


2

Exponential functions have the formula $$f(x)=a^{x-h}+k$$ Since you base is $e$, this then translates to $$f(x)=e^{x-h}+k$$ You have two points then. So plug them both in to get $$20=e^{1-h}+k$$ $$200=e^{4-h}+k$$ Thus $$20-e^{1-h}=200-e^{4-h}$$ $$e^{-h}(e^4-e)=180$$ $$e^{-h}=\frac{180}{e(e^3-1)}$$ $$-h=\ln{\frac{180}{e(e^3-1)}}$$ ...


3

Your logic is correct, and you have proved that the function is even. To draw a graph of the function, all you need to do is to draw the graph $y=|x|$ and shift it down by $3$ for the $-3$ factor in the equation. Here is the graph:


1

We know that $|x|=|-x|$, for all $x\in\Bbb R$. Therefore, $|x|-3=|-x|-3$, which is equivalent to $f(x)=f(-x)$. So yes, you are right. It is even.


0

The way you figure out whether it goes up or down in general is to pick points just next to the critical points of interest, and figure out if it is positive or negative, or a little more or a little less than its value at the critical point. For your example, let's figure out what it looks like at the far right of the diagram. Here we have passed all the ...


0

If the angle is found in any regular polygon, i.e., is of the type $$\frac{180^\circ(n-2)}n$$ then it will form an $n$-sided regular polygon, else it will form a spiral. This will intersect itself if the angle is less than $60^\circ$, and will form a circle as it approaches $180^\circ$.


3

You can use domain coloring: The two roots can be seen clearly: they are the points having a rainbow around it.


3

Yes, $F(x)=x$ for all $x\in[0,1]$. Sketch of proof: $\Phi(n)$ counts all primitive lattice points in the triangle with vertices $(0,0)$, $(0,n)$, and $(n,n)$. (Here a lattice point $(a,b)$ is primitive if $\gcd(a,b)=1$. The bijection sends $(a,b)$ to $a/b$.) You can count primitive lattice points by inclusion-exclusion: if $F(n)$ is the total number of ...


0

You need to convolve your data $\mathbf{x}$ with the impluse response of the corresponding FIR filter $\mathbf{h}$. You can learn the details from here.


0

Another way is, asking the plot of $g(x):=\frac{x-4}{|x-4|}\frac{1}{|x-4|}$ from maple. The result is: I saw this method in the textbook Thomas Calculus 11th ed


0

I think what you look for is the surd command. To plot $1/(x-4)^{1/3}$ in maple you can do like this: plot(surd((x-4),-3),x=-10..10); The result is In Mathematica you can do Plot[Surd[x - 4, -3], {x, -10, 20}] which results in



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