已知二次函数y＝x²＋ax＋b－3，x∈R的图像恒过点（2，0）则a²＋b²的最小值？, 已知二次函数y＝x²＋ax＋b
已知二次函数y＝x²＋ax＋b－3，x∈R的图像恒过点（2，0）则a²＋b²的最小值？
匿名 已知二次函数y＝x²＋ax＋b－3，x∈R的图像恒过点（2，0）则a²＋b²的最小值？
∵二次函数y=x平方+ax+b-3的图像恒过点（2，0）将(2,0)代入y=x²+ax+b-34+2a+b-3=0得b=-1-2a代入可得，a&#17罚锭窜瓜诃盖撮睡郸精8;+b²=a²+(1+2a)²=5a²+4a+1=5(a+2/5)²+1/5∴最小值为1/52√2R² =87990.3 求R等于多少_百度知道
2√2R² =87990.3 求R等于多少
因为A（6;﹢﹙2－b﹚²2;＝r&#178，圆半径为r＝﹙5√2﹚/2，-1&#47，2）C（0;＝r²（0－a﹚&#178，-3）在圆上;;﹢﹙y－b﹚²﹢﹙0－b﹚&#178，则该圆的标准方程为（x－a﹚&#178，b）;2);解这个三元一次方程组.圆心为C（5/2代入标准方程即得该圆的方程为（x－5/2﹚²2﹚²2，得a＝5/＝50/（0－a﹚²＝r&#178，半径为r;﹢﹙y+1&#47，故有（6－a﹚²﹢﹙—3－b﹚&#178，0），r＝﹙5√2﹚&#47，B（0，b＝-1/＝r²2;4设该圆的圆心为C（a
为您推荐：
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁∫∫(x^2 +y^2)dxdy，其中D为圆域(x-a)^2+y^2&=a^2,求详细解答在线等。_百度知道
∫∫(x^2 +y^2)dxdy，其中D为圆域(x-a)^2+y^2&=a^2,求详细解答在线等。
提问者采纳
//h.com/zhidao/wh%3D600%2C800/sign=a079cd3bd6f/314e251f95cad1c8f0c9abef7f3e3.baidu
来自团队：
其他类似问题
为您推荐：
其他3条回答
π)= ∫ [4a&#acosθ] dθ (θ; + y²&#8322；
θ ：-½π → +&#189:0→2acosθ]
dθ (θ;πa⁴ = a²;π→+½π) = ∫ [¼π→+½cos⁴θ] dθ (θ;π]= ∫ [ ∫ ρ³π)= &#179：-&#189：0 → 2a cos θ;dρdθ
[ρ ;,(ρ;dρ (ρ;&#47：ρ = 2a cos θ。此圆的极坐标方程是;π → +&#189，0)处，圆心在(a，这是一个圆方程;π→+&#189：-&#189：-&#189：-&#189，半径为a;ρ&#8308【解答】 (x - a)&#178， θ ;π直角坐标系中的面积元 dxdy 变成了 ρdρdθ 原积分= ∫∫ ρ&#179
最初答错了，隐藏回答。
看不懂你的问题啊。。。。。。
为什么。。。
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁Heart Curve
Heart Curve
Contents of this Page
What is the Heart Curve?
"Mathematische Basteleien"
What is the Heart Curve?
The heart curve is a closed curve, which has the shape
of a heart.&
The heart is well known as a figure on playings cards
besides diamonds, cross and spades.
If you speak about a heart, you rather mean the heart
figure than the heart shaped curve.&
In the simplest case, a heart is formed by a square standing
on its point and two semi-circles sitting on the sides. Characteristics
of the heart figure obviously are a groove above and a point below.&
A heart figure develops also if you set two semi-circles
upon a triangle. But here you get two unpleasant corners.
Obviously you expect that these sides are rounded.&
If the point below is missing then don't speak of a heart
but rather of a heart shaped& figure. This form, however, is more
similar to the human heart.&
The figure on the left is formed by three semi-circles.&
Drawn Heart Curves
1 Draw an isosceles triangle.
2 Draw the perpendiculars to the legs.&
They produce a second isosceles triangle.&
3 Draw two semi-circles upon the legs of the (now yellow)
triangle.&
1 Draw two touching equal circles.&
2 Draw the common tangent.
3 Draw two further (outer) tangents from one point of
the& tangent.
1 Draw a square.&
2 Draw four equal circles. The centres are the corners
of the square and the common radius is "half square side".&
1 Draw an ellipse.
2 Turn it about 45&.
3 Reflect it.
4 Form two hearts.&
1 Draw the graph of f(x)=sin(x), 0&x&pi/2.&
2 Turn the curve 90&.& Reflect this curve.&
3 Form a triangle of these two curves and a straight
4 Set two semi-circles upon the triangle.&
1 Draw the graph of f(x)=sin(x), -pi/2&x&pi/2.&
2 Turn the curve 90&. Reflect this curve.&
3 Form a triangle of these two curves and a straight
4 Set two semi-circles upon the triangle.&
Calculated
Heart Curves&&&&
It is a challenge to find formulas
which produce hearts.
You can describe method 4 by formulas.
The black ellipse has the formula 2x²-2xy+y²-1=0.The
domain is {x| x>=0}.
The red ellipse has the formula 2x²+2xy+y²-1=0.
The domain is {x| x&=0}.
You can leave out the domain, if you isolate y and use the
function f(x)=|x|.
Then y=|x|+sqrt(1-x²) and y=|x|-sqrt(1-x²)
describe a heart curve [(7)].
The formulas y=sqr(|x|)+sqrt(1-x²)
/\ y=sqrt(|x|)-sqrt(1-x²) are even better.
More curves
(Picture 1) Book 8, Eugen Beutel (1901) (2)
Buch 4, Aufgabe 8.5.5.,
(3) MathWorld (simplified formulas), (4)
H.-J. Caspar's web site&& (URL below), (6) Jurjen N.E.
From the desk of Torsten Sillke
Threedimensional
If you choose y=0 respectively x=0, you get the equation
of the 2D heart above on the left.&
Source: Gabriel Taubin [for example MathWorld (URL
The graphs were made by the
freeware program "winplot" (Version 23.05.2000, URL below).
Winword Hearts
And how do artists design a heart?&
The heart appears as a well known figure in character
sets of programs under MS Windows. There it is a figure of& playings
cards besides diamonds, cross and spades.&
Here is a choice of& well known character sets.
The sets are Normaler Text, Arial, Courier New, Estrangelo
Edessa, Lucida Console, Symbol, Times New Roman, Webdings.
If you increase the letters from 12 to 72 you recognize
the shapes.&
The originally black colour is replaced by the heart colour
The upper part of the heart figure is formed by curves
similar to arcs.& The lower lines do not approach linearly to the
point but usually are at first inward and then outward curved.& That
gives the heart a special sweep.
Found at Unicode Standard,
Version 4.0
1 BLACK HEART SUIT 2665&
2 WHITE HEART SUIT 2664
3 HEAVY BLACK HEART 2764
Pupils' Hearts
And how do students draw a heart?&
23 pupils of HS Lohfeld at Bad Salzuflen (Germany) were
to draw a simple heart.&
The results are:
Thanks to Klasse 7a,& Jg., about 12 years
Heart Curve or Cardioid
How to produce it
Draw a circle (on the left, yellow) and roll an equal
circle on it.&
Fix one point on the moving circle line and follow this
point. It describes the heart curve or cardioid (on the right).
way of producing a heart
A cardioid can also be seen as the envelope of circles.&
Draw a (yellow) circle and a fixed point P on the circle
All circles that pass through the fixed point P and have
their centres on the (yellow) circle have a cardioid as an envelope.
and perimeter of the heart curve
Use the polar form& r=2a[1+cos (t)] as the simplest
equation for calculating the area A and the perimeter U. The origin of
a coordinate system lies in the point of the cardioid.&
The perimeter is a rational number. A square with the
side 4a has the same one.
Mandelbrot
Set and Cardioid
The "main figure" of the Mandelbrot set has the form
of the cardioid. The German name Apfelm&nnchen (apple man) uses this
Actually the main figure is a cardioid. The points
of the Mandelbrot set, which have convergent sequences, lie inside a cardioid.&
Source: (5), page 208ff. There you find a proof and more
references.&
The picture was taken from my page .&
Catacaustic
and Cardioid
If light is falling on a spheric mirror (wedding ring
in the sun light), the reflecting rays form a special surface, the catacaustic.
It isn't a cardioid but a nephroid. A cardioid develops as an envelope,
if the rays start at a point on the circle and are reflected at the circle
(drawing on the right).&
You find more on my German page .
Characteristic
Curve of a Microphone
Microphones have a certain characteristic curve. In the
plane it is a circle for the "sound-pressure-receiver" and similar to a
lying eight figure for the "sound-velocity-receiver".
Special receivers like condenser microphones have both
capacities. Their characteristic curve develops by overlaying to a cardioid.&
Source: (6), page 550
The Broken Heart
The broken heart (Das gebrochene Herz) is a tangram game.
Lay inside a square two circles and draw some lines.
A heart develops, which is divided in nine pieces. The fun is to lay a
heart with the pieces or to discover new figures like those on the right.&
Source: (1) page 22, (2) page 140-145
The Woven Heart
1 Draw a square and two half circles sitting on it.
2 Cut it in along the red line.&
3 Copy it.& Paint the paper in two different colours
or start with coloured paper.&
4 Insert the blue piece in the green piece.&
5,6 The heart works fine with a larger number of slits
If you like have a look at the
page made by Christopher Hamkins.
Tesselation of
1 Give a spiral.
2 Reflect the spiral on the end point.
3 Connect both spirals to get a double spiral.
4 Reflect the double spiral. It forms a heart together
with the first double spiral.
Many hearts lead to a tesselation (on the right).
The idea for this drawing came from a window grill in Venice
(June 2004):&
Venice is rich with heart
Rosettes of
Heart Figures
four-leaf clover
Once again a Venice photo- brightened up with some red.&
Six waffles
A pair of swans during the courtship display
Photo on the web site fotocommunity (URL below)
Emoticons&&
References&&
(1) Pieter van Delft, Jack Botermans: Denkspiele der
Welt, M&nchen 1998 ISBN 3-]&
(2) Karl-Heinz Koch: ...lege Spiele, K&ln 1987 (dumont
taschenbuch1480)& [ISBN 3-]&
(3) Heinz Nickel u.a: Algebra und Geometrie f&r Ingenieur-
und Fachschulen, Frankfurt / Z&rich 1966
(4) Hans Schupp, Heinz Dabrock: H&here Kurven,&
BI Wissenschaftsverlag 1995 [ISBN 3-411-17221-5]&
x^2 + 2( 3/5 (x^2)^(1/3) - y )^2
(5) Herbert Zeitler: &Uber die Hauptk&rper spezieller
Funktionen, MNU, Jg.52, 1999, Heft4
(6) Bergmann-Schaefer: Lehrbuch der Expermentalphysik,
Berlin, NewYork 1975 [ISBN 3 11 ]
(7) Norbert Herrmann: Mathematik ist &berall, Oldenbourg
Verlag 2004 [ISBN 3-486-57583-X]
&y = |x| +- sqrt(1 - x^2)
y = 2/3 ( (x^2 + |x| - 6)/(x^2
+ |x| + 2) +- sqrt(36 - x^2) ) (siehe auch Webseite von Thomas Jahre)&
(8) Eugen Beutel: Algebraische Kurven, G.J. G&schen,
Leipzig 1909-11
(x^2 + y^2 - 1)^3 = 4x^2y^3
(9) Ulrich Graf: Kabarett der Mathematik,& Dresden:
L. Ehlermann, 1942 Hardcover, 1.Auflage. (1943 Hardcover. 2. Auflage.)
(10) Michael Zettler: Und noch ein Herz. PM 6/99 Seite
y = sqrt(1 - (|x|-1)^2),&
y = arccos(1 - |x|) - pi
(11) Thomas Hechinger:&& ... und noch ein weiteres
Herz. PM 2/00 Seite 67
y = sqrt(1 - (|x|-1)^2),&
y = -3 sqrt(1 - sqrt(|x|/2))
(12) Mitteilung von& Torsten Sillke:&
x^2 + 2 (y - p*|x|^q)^2 = 1 (siehe
Schupp / Eisemann)
r = 2 sin^2(phi/4) = 1 - cos(phi/2)
mit |phi| &= pi& (nach Eisemann)
r = |phi|/pi&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
mit |phi| &= pi. (Archimedische Spirale)
r = (1 - |phi|)(1 + 3|phi|)&&&&&&&&&&&&&&&
mit |phi| &= 1. (nach Caspar)
(13) El-Milick, Elements d'Algebre
Ornementale, Paris, 1936:
y=(x)^(2/3)+(a²-x²)^(1/2)
und y=x^(2/3)-(a²-x²)^(1/2) and a=2
Heart Curves
on the Internet
33. Evangelischer Kirchentag in Dresden 2011
Armin Dietz
Christian Ucke und Christian Engelhardt
[erschienen in: Physik in unserer Zeit, 29 (1998), Seite
120 bis 122]
Benutzer:Georg-Johann (Wikipedia)
x(t)=12sin(t)-4sin(3t)
y(t)=13cos(t)-5cos(2t)-2cos(3t)-cos(4t)
Ernst-Moritz-Arndt-Universit&t Greifswald
(Mathematik und Kunst)
Dies ist die Formel des Randes des
"Herzkartenentwurf von Stabius-Werner"
X =&&&&&&& t sin(
3.14 sin(t)/t )
Y = - abs(t) cos( 3.14 sin(t)/t )
-pi &= t &= pi
(Mitteilung von Torsten Sillke)
Fotocommunity
(Thomas Th.)&
Hans-J&rgen (Matroids Matheplanet)
Hans-J&rgen Caspar
x = a (-phi² + 40 phi +1200)
sin(pi*phi/180)
y = a (-phi² + 40 phi +1200)
cos(pi*phi/180)
Jan Schormann
Friedrich Krause
y = sqrt(|x|) +- sqrt(1 - x^2)
Klaus Rohwer
Michael Holzapfel
y = sqrt(1 - (|x|-1)^2),&
y = arccos(1 - |x|) - pi&
(x^2 + y^2)(x^2 + y^2 - 2ax) -
a^2y^2& =& 0 (Kardioide)
NN (Matheplanet)
Thomas Jahre (Chemnitzer Schulmodell)
Ein Herz f&r die Mathematik
Torsten Sillke
Wikipedia&
Alex Bogomolny (Cut The Knot!)
Eric W. Weisstein (MathWorld)
(x²+y²-1)³-x²y³
(siehe Beutel)
x=sin(t)cos(t)ln|t| ; y=|t|0.3[cos(t)]0.5
mit 0&=t&=1
[x²+(9/4)y²+z²]³-x²z³-(8/90)y²z³=0
(2x²+2y²+z²-1)³-(1/10)x²z³-y²z³=0
Jan Wassenaar
JOC/EFR (School of Mathematics and Statistics, University
of St Andrews, Scotland)
Kurt Eisemann
x^2 + (y - 3/4 (x^2)^(1/3))^2 =
1 (Footnote)
r = sin^2( pi/8 - phi/4 ) (Footnote)
<font size=+
Pavel Boytchev&
&& (Video)
Richard Parris (peanut Software)&
Robert FERR&EOL (mathcurve)
y=cos t -sin4 t
Nobuo YAMAMOTO
Thanks to Torsten Sillke for several hints.
Gail from Oregon Coast - thank you for supporting me
in my translation.
Found on the Internet
the nerdy way of drawing a heart.
http://www.mathematische-basteleien.de/heart.htm
Must they do EVERYTHING in math? >&;; lol.
Feedback: Email address on my main page&
This page is also available in .&
URL of my Homepage:
J&rgen K&ller 2004已知抛物线y=x&#178;+3x-2如图所示,那么点（b,c）在哪一象限?错了，题目应该是已知抛物线&y=x&#178;+bx+c&如图所示，那么点（b,c）在哪一象限？
阿K第五季28b
你是不是想问若y=ax^2+bx+c的图像如图所示,则点（b,c)在哪一象限?由图可知,a＞0,c＜0.且-b/2a＜0,所以b＞0.因此（b,c)在第四象限 如果不是,那我实在不明白你要问什么
你太聪明了，我打错题了。
由图如何得知 c＜0的呢？
对于y=ax^2+bx+c，当x=0时，y=c，因此c就是当x=0时的函数值，也就是函数图象与y轴的交点，由图像可知，函数图像与y轴的交点在x轴的下方，所以c＜0
为您推荐：
其他类似问题
图上没标b.c是哪里啊
很容易,由图知,对称轴-b/2a0,所以b>0.而c为抛物线与y轴交点坐标,c<0, 点（b,c）在第四象限
扫描下载二维码