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## Analyze and graph line equations and functions step-by-step

\bold{\mathrm{Basic}}
\bold{\alpha\beta\gamma}
\bold{\mathrm{AB\Gamma}}
\bold{\sin\cos}
\bold{\ge\div\rightarrow}
\bold{\overline{x}\space\mathbb{C}\forall}
\bold{\sum\space\int\space\product}
\bold{\begin{pmatrix}\square&\square\\square&\square\end{pmatrix}}
\bold{H_{2}O}

\square^{2}
x^{\square}
\sqrt{\square}
\nthroot[\msquare]{\square}
\frac{\msquare}{\msquare}
\log_{\msquare}
\pi
\theta
\infty
\int
\frac{d}{dx}
\ge
\le
\cdot
\div
x^{\circ}
(\square)
|\square|
(f\:\circ\:g)
f(x)
\ln
e^{\square}
\left(\square\right)^{‘}
\frac{\partial}{\partial x}
\int_{\msquare}^{\msquare}
\lim
\sum
\sin
\cos
\tan
\cot
\csc
\sec
\alpha
\beta
\gamma
\delta
\zeta
\eta
\theta
\iota
\kappa
\lambda
\mu
\nu
\xi
\pi
\rho
\sigma
\tau
\upsilon
\phi
\chi
\psi
\omega
A
B
\Gamma
\Delta
E
Z
H
\Theta
K
\Lambda
M
N
\Xi
\Pi
P
\Sigma
T
\Upsilon
\Phi
X
\Psi
\Omega
\sin
\cos
\tan
\cot
\sec
\csc
\sinh
\cosh
\tanh
\coth
\sech
\arcsin
\arccos
\arctan
\arccot
\arcsec
\arccsc
\arcsinh
\arccosh
\arctanh
\arccoth
\arcsech
+

=
\div
/
\cdot
\times
<
” >>
\le
\ge
(\square)
[\square]
▭\:\longdivision{▭}
\times \twostack{▭}{▭}
+ \twostack{▭}{▭}
– \twostack{▭}{▭}
\square!
x^{\circ}
\rightarrow
\lfloor\square\rfloor
\lceil\square\rceil
\overline{\square}
\vec{\square}
\in
\forall
\notin
\exist
\mathbb{R}
\mathbb{C}
\mathbb{N}
\mathbb{Z}
\emptyset
\vee
\wedge
\neg
\oplus
\cap
\cup
\square^{c}
\subset
\subsete
\superset
\supersete
\int
\int\int
\int\int\int
\int_{\square}^{\square}
\int_{\square}^{\square}\int_{\square}^{\square}
\int_{\square}^{\square}\int_{\square}^{\square}\int_{\square}^{\square}
\sum
\prod
\lim
\lim _{x\to \infty }
\lim _{x\to 0+}
\lim _{x\to 0-}
\frac{d}{dx}
\frac{d^2}{dx^2}
\left(\square\right)^{‘}
\left(\square\right)^{”}
\frac{\partial}{\partial x}
(2\times2)
(2\times3)
(3\times3)
(3\times2)
(4\times2)
(4\times3)
(4\times4)
(3\times4)
(2\times4)
(5\times5)

(1\times2)
(1\times3)
(1\times4)
(1\times5)
(1\times6)
(2\times1)
(3\times1)
(4\times1)
(5\times1)
(6\times1)
(7\times1)
\mathrm{Degrees}
\square!
(
)
%
\mathrm{clear}
\arcsin
\sin
\sqrt{\square}
7
8
9
\div
\arccos
\cos
\ln
4
5
6
\times
\arctan
\tan
\log
1
2
3

\pi
e
x^{\square}
0
.
\bold{=}
+

## Most Used Actions

\mathrm{simplify}

\mathrm{solve\:for}

\mathrm{inverse}

\mathrm{tangent}

\mathrm{line}

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Steps for Sketching the Graph of the Function on eMathHelp – If function is obtained by transforming simpler function, perform corresponding shift, compressing/stretching. It is often convenient to draw all points Example 3. Sketch graph of the function f(x)=(x^2-5x+6)/(x^2+1). There is no simpler function that initial function is obtained from.The graph of a quadratic function is a parabola. The parabola can either be in "legs up" or "legs down" orientation. We know that a quadratic equation will be in the form If you are trying to find the zeros for the function (that is find x when f(x) = 0), then that is simply done using quadratic equation – no…Since in our problem the function g(x) = f(x – c) + k, we have shifted the function f(x) c units to the right and k units upward, that is, we have shifted the function f(x) 5 You know the right answer? The graph of f(x) = x2 is translated to form g(x) = (x – 5)2 + 1.which graph represents g(x)?…

How to find the equation of a quadratic function from its graph – Start studying Quadratic Functions Unit Test. Learn vocabulary, terms and more with flashcards, games and other study tools. What value represents the vertical translation from the graph of the parent function f(x) = x2 to the graph of the function g(x) = (x + 5)2 + 3?Graph a Linear Function f(x) = 3x – 2. 131 326 просмотров 131 тыс. просмотров. • 22 мая 2012 г. How to graph a linear function using the slope and y-intercept.calculista calculista. we have. This is a quadratic equation (vertical parabola) open up. The vertex is a minimum-> vertex is the point. The domain of the function is all real numbers-> interval (-∞,∞).

The graph of f(x) = x2 is translated to form g(x) = (x – 5) 2 + 1.which… – 8. Find The Critical Points For The Function F(x) – 12x-x. Transcribed Image Text from this Question. 7. Which graph represents the function f(x) = x(x – 1)(x-3)?An easy way to think about discontinuity is that a discontinuity exists whenever you cannot draw that part of the graph without lifting up your pencil. If we look at the parts of the function, f(x), it is the sum of continuous functions, all of which are defined for all real numbers.Step-by-step explanation: Given : Function. To find : Which graph represents ? Solution The vertex of the upward parabola is (-2,-3). The x-intercept are (-3.732,0) and (-0.268,0).

Vector function for the curve of intersection of two surfaces (KristaKingMath) – .

Solving Rational Inequalities – .

Closest Point to the Origin | MIT 18.01SC Single Variable Calculus, Fall 2010 – CHRISTINE BREINER: Welcome
back to recitation.
Today we're going to work
on an optimization problem. So the question I
want us to answer is, what point on the curve y
equals square root of x plus 4 comes closest to the origin? I've drawn a sketch
of this curve. The scale in this
direction– each hash mark is one unit in the x
direction, each hash mark here is one unit in the y direction. Just want to point
out two easy places to figure out the
distance to the origin. Over here, where the curve
starts at negative 4, 0, the distance to the
origin is 4 units. And here at (0, 2) the distance
to the origin is two units. It's probably, we could
safely say, further away here. So we're anticipating
that somewhere along the curve
in this region is where we should find our place
that's closest to the origin. The only reason I
point that out is that, when you're doing these problems
on your own you should always try and anticipate roughly
where things should happen, in what kind of region, so that
you don't– you don't start thinking, if you do something
wrong and you get x equals 100 and then you come back
and look at the curve, you realize right away, well,
that doesn't make any sense. So we want to always be thinking
as we're solving the problems, does my answer make sense? So I'm actually going to
give you a little bit of time to work on this yourself
and then I'll come back and I'll work on it as well. Welcome back. Hopefully you were able to get
pretty far into this problem. And so I will start
working on it now. So again, the
question is that we want to optimize– in this
case, minimize– the distance to the origin from this curve. And so what we're really trying
to do is we have a constraint, the constraint is we
have to be on the curve, and then we also have something
we're trying to minimize. And the thing we're trying
to minimize is distance. And so we have to make sure that
we understand the two equations that we need– the optimization,
or the constraint equation, and the optimizing equation. So to optimize we need to
know how to measure distance in two-dimensional space. And one point I want
to make is that if you want to optimize distance
you might as well optimize the square of distance
because it's much easier. So let me justify that
briefly and then we'll go on. So I want to optimize the
distance squared to the origin. It's, well distance, you
remember, first in general, between two points
(x, y) and (a, b) is something in this form. Distance squared
is the difference between the x-value
squared plus the difference between the y-value squared. This is, should remind you
of the Pythagorean theorem, ultimately. So in this case, in our
case, distance to the origin is x squared plus y squared. The distance squared is
x squared plus y squared. I just told you that instead
of trying to optimize distance, we can optimize
distance squared. Why is that? Well, remember that
when you optimize, what you're looking
for is a place where the derivative of the function
of interest is equal to 0. So what I want to
point out is that when you take the
derivative of distance squared and find
where that's 0, it's going to be the
same as the place where the derivative of
distance is equal to 0. So let's notice that. So this is a little
sidebar justification. Notice d squared prime
is equal to 2d d prime. Where did that come from? That's this is implicit
differentiation with respect to x and
this is the chain rule. So if I want d
prime to equal 0, I can also find where d
squared prime equals 0. I'm assuming– notice
the distance is never at the origin– so
distance is never 0. So I don't have to
worry about that. So that's a small
sidebar, but just to justify why we can do that. Now let's come back into
the problem at hand. What is our optimization
problem, equation that we want to minimize? We want to minimize this
equation with respect to a certain constraint. What's the constraint? The constraint is what y is. y depends on x. And so when I solve
these problems I'm going to have to
substitute in my constraint. So y squared is the square root
of x plus 4 quantity squared, so I just get x plus 4. So now I have my
optimization equation. How do I find a
minimum or a maximum? I take the derivative
and set it equal to 0. So let me come give
myself a little more room and do that over here. So d squared prime, now I get
derivative of x squared is 2x. The derivative of x is 1,
and the derivative of 4 is 0. This will be optimized
when this is equal to 0. So 0 equals 2x plus 1. So x is equal to minus 1/2. Does this pass, as we would
say maybe, the smell test? Does it smell OK to us? The answer will be yes. Because remember, we said
somewhere in this x region is where we expect that we
will have a distance closest, point closest to the origin. And so we're right
here on the x value. Now we have to find what the y
value is to finish the problem. But this is not, so
far, very surprising. It seems like maybe
the right thing. Now we have x. So now how do we find y? Well, we know what y is. y is equal to the
square root of x plus 4, so it's equal to the square
root of negative 1/2 plus 4, which simplified is 3 and
1/2, which I think is 7/2. So the point is negative 1/2
comma square root of 7/2. And then you just want to
double check and make sure, did I ask for the distance
or did I ask for the point? And right now we have the
point, so let's come over and make sure what
point on the curve comes closest to the origin. So now we know that we've
answered the correct question. So again, it was a
maximize– sorry, it was a minimizing problem. It was an optimization
problem where we wanted to minimize distance. We had a constraint equation. We had the thing we
wanted to minimize. And then we took the
derivative of the minimizer, set it– of the optimizing
equation, set it equal to 0, solved for x, and then found the
answer to the specific question by then finding the y-value. And I think I'll stop there. .