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Usually, a compound microscope comes with 3 or 4 objective lenses. The most common setting is: Scanning objective lens (4x) A scanning objective lens provides the lowest magnification power of all objective lenses. The name 'scanning' objective lens comes from the fact that they provide observers with enough magnification for a wide. PicFocus 2.3 Moon Hunters (2016) 3D Converter 6.5.9 Crack FL Studio Producer Edition 12.4.29 Aimersoft DVD Creator 6.0.1 Crack Musictube 1.8 Crack 4Video DVD Manager 5.2.31 Matchbook 1.1.1 Shimo 4.1.5.1 CleanMyMac 3 V3.9.0 Full Crack Rocket Pro 1.5 Quixel Mixer 2018.2.4.0 Crack Mac Free (FULL). Once you have proceeded up to Step 2 of Part I, set the lens focus mode switch to. To shoot still photos, make sure that the camera is ready to take shots, and then press the (Live View shooting) button. To shoot movies, set the Mode Dial to. The Live View image will appear on the LCD monitor.

Objectives:

• Lesson 1: Find the standard form of a quadratic function, and then find the vertex, line of symmetry, and maximum or minimum value for the defined quadratic function.
• Lesson 2: Find the vertex, focus, and directrix, and draw a graph of a parabola, given its equation.
• Lesson 3: Find the equation of our parabola when we are given the coordinates of its focus and vertex.
• Lesson 4: Find the vertex, focus, and directrix, and graph a parabola by first completing the square.

Lesson 1

The Parabola is defined as 'the set of all points P in a plane equidistantfrom a fixed line and a fixed point in the plane.' The fixed line iscalled the directrix, and the fixed point is called the focus.

A parabola, as shown on the cables of the GoldenGate Bridge (below), can be seen in many different forms. The paththat a thrown ball takes or the flow of water from a hose each illustratethe shape of the parabola.

Each parabola is, in some form, a graph of a second-degree function andhas many properties that are worthy of examination. Let's begin by lookingat the standard form for the equation of a parabola.

The standard form is (x - h)² = 4p (y - k), where the focusis (h, k + p) and the directrix is y = k - p. If the parabola is rotatedso that its vertex is (h,k) and its axis of symmetry is parallel to thex-axis, it has an equation of (y - k)² = 4p (x - h), where thefocus is (h + p, k) and the directrix is x = h - p.

It would also be in our best interest to cover another form that theequation of a parabola may appear as
y = (x - h)² + k, where h represents the distance that the parabolahas been translated along the x axis, and k represents the distance theparabola has been shifted up and down the y-axis.

Mainstage 3 3 3 1. Completing the square to get the standard form of aparabola.

We should now determine how we will arrive at an equation in the formy = (x - h)² + k;

Example 1

Suppose we are given an equation like
y = 3x² + 12x + 1.

We now need to complete the square for this equation. I will assume thatyou have had some instruction on completing the square; but in case youhaven't, I will go through one example and leave the rest to the reader.

When completing the square, we first have to isolate the Ax²term and the By term from the C term. So the first couple of steps willonly deal with the first two parts of the trinomial.

In order to complete the square, the quadratic in the form y = Ax²+ By + C cannot have an A term that is anything other than 1. In our example,A = 3; so we now need to divide the 3 out, but that is only out of the 3x²+ 12x terms.

Picfocus 2 3d Printer

This simplifies to y = 3(x² + 4x) + 1. From here we need totake 1/2 of our B term, then square the product. So in this case, we have1/2(4) = 2, then 2² is 4. Now, take that 4 and place it insidethe parenthetical term.

To update what we have: y = 3(x² + 4x + 4) + 1; but we nowneed to keep in mind that we have added a term to our equation that mustbe accounted for. By adding 4 to the inside of the parenthesis, we havedone more than just add 4 to the equation. We have now added 4 times the3 that is sitting in front of the parenthetical term. So, really we areadding 12 to the equation, and we must now offset that on the same sideof the equation. We will now offset by subtracting 12 from that 1 we leftoff to the right hand side.

To update: y = 3(x² + 4x + 4) + 1 - 12. We have now successfullycompleted the square. Now we need to get this into more friendly terms.The inside of the parenthesis (the completed square) can be simplified to(x + 2)². The final version after the smoke clears is y = 3(x+ 2)² - 11. And , oh the wealth of information we can pull fromsomething like this! We will find the specifics from this type of equationbelow.

Finding the vertex, lineof symmetry, and maximum and minimum value for the defined quadratic function.

Let's first focus on the second form mentioned, y =(x - h)²+ k. When we have an equation in this form, we can safely say that the 'h'represents the same thing that 'h' represented in the first standard formthat we mentioned, as does the 'k'. When we have an equation like y = (x- 3)² + 4, we see that the graph has been shifted 3 units tothe right and 4 units upward. The picture below shows this parabola in thefirst quadrant.

Had the inside of the parenthesis in the example equation read,'(x+3)'as opposed to '(x-3),' then the graph would have been shiftedthree units to the left of the origin. The '+4' at the end ofthe equation tells the graph to shift up four units. Likewise, had the equationread '-4,' then the graph would still be pointed upward, but thevertex would have been four units below the x-axis.

A great deal can be determined by an equation in this form.

The Vertex

The most obvious thing that we can tell, without having to look at thegraph, is the origin. The origin can be found by pairing the h value withthe k value, to give the coordinate (h, k). The most obvious mistake thatcan arise from this is by taking the wrong sign of the 'h.' In our exampleequation, y = (x - 3)² + 4, we noticed that the 'h' is 3, butit is often mistaken that the x-coordinate of our vertex is -3; this isnot the case because our standard form for the equation is y = (x - h)²+ k, implying that the we need to change the sign of what is inside theparenthesis.

The Line of Symmetry

To find the line of symmetry of a parabola in this form, we need to rememberthat we are only dealing with parabolas that are pointed up or down in nature.With this in mind, the line of symmetry (also known as the axis of symmetry)is the line that splits the parabola into two separate branches that mirroreach other. The line of symmetry goes through the vertex, and since we arenow only dealing with parabolas that go up and down, the line of symmetrymust be a vertical line that will begin with 'x = _ '. The numberthat goes in this blank will be the x-coordinate of the vertex. For example,when we looked at y = (x - 3)² + 4, the x-coordinate of the vertexis going be 3; so the equation for the line of symmetry is x = 3.

In order to visualize the line of symmetry, take the picture of the parabolaabove and draw an imaginary vertical line through the vertex. If you wereto take the equation of that vertical line, you would notice that the lineis going through the x-axis at x = 3. An easy mistake that students oftenmake is that they say that the line of symmetry is y = 3 since the lineis vertical. We must keep in mind that the equations for vertical and horizontallines are the reverse of what you expect them to be. We always say thatvertical means 'up and down; so the equation of the line (being parallelto the y-axis) begins with 'y =__',' but we forget that the key iswhich axis the line goes through. So since the line goes through the x-axis,the equation for this vertical line must be x = __.

The Maximum or Minimum

In the line of symmetry discussion, we dealt with the x-coordinate ofthe vertex; and just like clockwork, we need to now examine the y-coordinate.The y-coordinate of the vertex tells us how high or how low the parabolasits.

Once again with our trusty example, y = (x-3)² + 4, we seethat the y-coordinate of the vertex (as derived from the number on the farright of the equation) dictates how high or low on the coordinate planethat the parabola sits. This parabola is resting on the line y = 4 (seeline of symmetry for why the equation is y = __, instead of x = __). Once we have identified what the y-coordinate is, the last question wehave is whether this number represents a maximum or minimum. We call thisnumber a maximum if the parabola is facing downward (the vertex representsthe highest point on the parabola), and we can call it a minimum if theparabola is facing upward (the vertex represents the lowest point on theparabola).

Omnigraffle pro 7 8 2 download free. How do we tell if the parabola is pointed upward or downward by justlooking at the equation?

As long as we have the equation in the form derived from the completingthe square step, we look and see if there is a negative sign in front ofthe parenthetical term. If the equation comes in the form of y = - (x -h)² + k, the negative in front of the parenthesis tells us thatthe parabola is pointed downward (as illustrated in the picture below).If there is no negative sign in front, then the parabola faces upward.

Example 2

Let's now look at an example of another equation of a parabola in standardform. We will then identify its vertex, line of symmetry, and maximum orminimum.

For example, let's take the equation y = - (x + 4)² - 7. Thefirst thing we would like to do is look at the graph of the curve. Thisshould help us make sense of the things we are looking for. The graph isshown below.

As you can see, this curve falls into the third quadrant and is pointeddownward. The vertex appears to have a negative x-coordinate and a negativey-coordinate. We will look more closely at the equation and take what wehave already learned, we should be satisfied with our results.

y = - (x + 4)² - 7 gives us a vertex of (-4, -7). The x-coordinateis the 'opposite' of 4, which is -4, and the y-coordinate is -7as seen from the number that sits at the end of the equation.

First, the negative sign at the beginning of the equation immediatelytips us off that the parabola is facing downward.

Next, the x-coordinate that we found is the key to finding the line ofsymmetry. We know that the equation for the line of symmetry will be 'x= __ ,' and the number inside the blank is the x-coordinate, -4.

Lastly, we need to decide whether we have a maximum or minimum. The y-coordinateis going to be a maximum in this case because the vertex lies on the highestpoint (maximum) of our curve. So in this case, we have a maximum of y =-7.

Let's now look at the same curve above with the vertex, line of symmetry,and maximum visible:

Obviously, the red curve represents the parabola. The green line representsour line of symmetry (equation x = -4) and the blue line represents theline that the maximum rests on at y = -7. Hopefully this visual has helpedyou see all of the specific parts that we have discussed so far.

For some supplementary exercises over what we have covered so far clickhere: Exercise1

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Lesson 2

Find the vertex, focus, and directrix, and draw a graphof a parabola, given its equation.

As you may or may not know, a parabola is the locus of points in a planeequidistant from a fixed line and a fixed point on the plane. We know thisfixed line to be the directrix and the fixed point to be the focus.

To see an animated picture of the above description, you need to haveGeometer's SketchPad for either Macintosh or PC loaded on your computer.If you have GSP, click here. To download thescript of this picture so you can create it yourself, clickhere.

Let's now take a look at a parabola that has all of the elements thatwe will be looking for:

• the vertex
• the focus
• the directrix

The following example is especially meant for those who do not have GSPon your computer. This picture (below) is generated from Algebra Xpresser.

From the above picture, I have labeled three items that we need to payclose attention to. The highest point of the parabola is the vertex (andthe maximum). The plus sign that is directly under the vertex is the focus.The green line that is above the parabola (and directly above the vertex)is the directrix. You may be able to see, by eyeballing, that the distancefrom the focus to the vertex is the same distance as the vertex to the directrix.We will now go into a bit of detail as to how to derive all of this informationfrom a given equation.

The next example that I will give you will be a nice, easy equation fromwhich we can easily pick the information we need.

Example 3

Let's examine the equation, (x + 2)² = -6(y - 1).

Obviously, this equation is different from the 'vertex form'we learned in the prior lesson. Even still we can find all of the informationwe found in the first lesson: the vertex, line of symmetry, and the maximum/minimum.We have to apply the same line of thought that the vertex is where the xand y terms are. In the same manner we find the x-coordinate is -2, they-coordinate is 1 {V: (-2, 1)}. All we need to find the line of symmetryand the maximum/minimum is the vertex; so let's follow through: The lineof symmetry is x = -2, and the maximum (since we have a negative sign infront of one of our terms) is at y = 1.

Now for the fun part.

In order to find the focus and directrix of the parabola, we need tohave the equations that give an up or down facing parabola in the form (x- h)² = 4p(y - k) form. In other words, we need to have the x²term isolated from the rest of the equation. We are used to having x²by itself, but if the vertex has been shifted either up or down, we needto show this in the parenthetical term with the y. The coefficient of the(y - k) term is the 4p term. We need to take this number and set it equalto 4p.

In this case, 4p is equal to the term in front of the y term (in parenthesis);so 4p = -6. This means that p = -3/2. Since this is an downward facing parabola,we need to have the focus inside of the curve, meaning the focus is belowthe vertex. How far below the vertex? Take the y-coordinate and add thep term it. So, we now have the vertex at (-2, 1) and we are, in essence,subtracting -3/2 from 1. This will move the focus to the point (-2, -1/2).

The directrix is equidistant from the vertex that the focus is. So ifthe focus is down -3/2 from the vertex, then the directrix is a line thatis up 3/2 from the vertex. That puts the directrix at y = 5/2.

For an illustration of this problem, please look at the picture below:

The green line (3/2 units up from the vertex) is the directrix, and theplus sign 3/2 units down from the vertex is the focus.

This should help us with the parabolas that open upward and downward.Let's now take a look at a parabolat that opens left and right.

Example 4

Let's take a look at the equation (y + 3)²= 12 (x - 1).

We can easily identify that the parabola is opening left or right. Sincethe coefficient in front of the x term is positive, we can say that theparabola will open to the right. The focus will be to the right of the vertex,and the directrix will be a vertical line that is the same distance to theleft of the vertex that the focus is to the right.

The vertex is (1, -3), the axis of symmetry (now horizontal) is y = -3,and we don't recognize 'max's and min's' for parabolas that openleft or right.

The term in front of the x term is a 12. This is what our 4p term isequal to. So 4p = 12, making p = 3. So we now need to move the focus 3 unitsright from the the origin. This means that the coordinate for the focusis (4, -3), and the directrix will be a vertical line going through thepoint (-2, -3).

This problem is illustrated in the picture below.

Once again, our green line represents the directrix and the plus signrepresents the aforementioned focus.

For some supplementary exercises over what we have covered in lesson2, click here: Exercise2

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Lesson 3

Find the equation of a parabola when we are given thecoordinates of its focus and vertex.

Now, we are going to begin taking what we have learned and start piecingit together. If we are given a focus and a vertex, we have enough to beable to generate a quadratic equation of a parabola. If we think about itfor a second, we will be able to find the distance from the vertex to thefocus based on this given information. We will then be able to calculateour p term (the term from the previous lesson that is in front of our non-squaredvariable). Placing the coordinates of the vertex into the equation is verysimple, relative to what we have learned so far.

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Example 5

Let us suppose that we are given a focus of (-6, 0) and the vertex isat the origin.

Based on what we know without plugging anything in, we can say that theparabola will be opening up to the left because its focus is to the leftof the origin. Now in beginning to piece things together, we can say thatthe equation will be something like y² is equal to some x term.

Grabit 4 920 hp. Since the origin is the vertex, we can say that this will be (y - 0)²= 4p(x - 0), which simplifies to y² = 4px.

We know that p = -6, and we know that 4p = -24. We should now be ableto tell that the equation is y² = -24x.

Example 6

We will now try a problem that has the parabola opening up or down. Wewill make the focus (2, 3) and the vertex (2, 6).

The focus is directly below the vertex by 3 units; so p = -3; so 4p =-12; but not so fast! We aren't quite home free yet. The vertex is shiftedoff of the origin, and we need to consider the h and k terms.

The equation with a parabola facing downward will be (x - h)²= 4p(y - k), where 4p is negative. To again piece things together:
(x - 2)² = -12 (y - 2).

For some supplementary exercises over what we have covered in lesson3, click here: Exercise3

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Lesson 4

Find the vertex, focus, and directrix, and graph a parabolaby first completing the square.

Not always do we come up on equations that are there just waiting forus to solve them. Sometimes we've got to work a bit to find their key points.Hopefully this example will lead us toward such a problem.

Example 7

The last pair of examples that we will examine will be one where we aregiven a quadratic equation that is not already in any particular standardform.
We will now be forced to complete the square to arrive at the form we needto find the newest parts of the parabola that we have explored.

Suppose that we have x² + 6x +4y + 5 = 0. Since the x-termis the squared term, we will choose to isolate all of the terms that havex in them. We will need to place the x terms on one side of the equation,while the rest of the terms are on the opposite side.

This step will leave us with x² + 6x = -4y - 5. When we completethe square on the left-hand side of the equation, we will have x²+ 6x + 9; sowe will need to add 9 to the right-hand side, as well.

This will bring us to (x + 3)² = -4y + 4. Remembering thatany coefficients of the x or y terms need to go in front of the non-squaredvariable, we will factor the -4 from the y-term. This will leave us with(x + 3)² = -4 (y - 1).

From here, the problem resembles both of the others.

Our vertex is (-3, 1), our line of symmetry is x = -3; and we do havea maximum at y = 1;

The focus can now be found by taking the number in front of the non-squaredvariable -4 and setting it equal to 4p. 4p = -4; so p = -1.

Since the parabola is facing downward, the focus is below the vertex,and the directrix is above. We will take our vertex and add (-1) to they-coordinate. This will take us to the point (-3, 0) that is our focus.The directrix (on the opposite side of the vertex) is at the horizontalline y = 2. Once again, we will look at an illustration below. The greenline is the directrix, and blue dot is the focus.

(Corrections by J. Wilson, 28 Feb 2012)

Example 8

Suppose that we have an equation y² + 2y + 4x -8 = 0. Beforewe doing any steps, such as completing the square, we can see that the squaredterm is the y term. This will tell us that the parabola with either opento the left or to the right. Since this is the case where the y term issquared, we need to isolate the y terms to one side of the equation andput the x terms and the constants on the other. We will stick to what we'vebeen doing all along and put the isolated terms on the left.

After the isolation step, we see that we have y² + 2y = -4x+ 8.

In completing the square, we will have y + 2y + 1 = -4x + 8 + 1; thissimplifies to (y + 1)² = -4x + 9.

No matter how ugly the right-hand side of the equation may get, we needto divide the right hand side by the coefficient of the x term (in thiscase, -4). This will leave us with (y + 1)² = -4(x - 9/4). Fromhere we can say that the parabola will open to the left.

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We can now see that the vertex will be at (9/4, -1).

The term in front of the x is a -4. This is our 4p value. So we now cansay that 4p = -4. In turn, our p = -1.

Now will we determine that the focus is one unit left of the vertex;so the focus (after some work with fractions) is (5/4, -1).

The directrix is going to be a vertical line that is one unit to theright of the vertex. So the directrix will be a line, x = 13/4.

Below, we will see the sketch of this equation.

As has been the case so far, the plus sign represents the focus, whichresides at the point (5/4, -1); the directrix is represented by the greenline, which is on the equation x = 13/4.

For some supplementary exercises over what we have covered in lesson4, click here: Exercise4