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# how do I draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), and (5,5)

draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), and (5,5)

## Research, Knowledge and Information :

### how do I draw cubic function as a dashed curve using the ...

draw cubic function as a dashed curve using the points (-4,3),(0,0), ... how do I draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), ...

### Graphing Cubic Functions - Free Mathematics Tutorials ...

Graphing Cubic Functions. ... The x intercept are at the points (0 , 0). b ... 2) 3 = 0 Function f has one zero at x = 2 of multiplicity 3 and therefore the graph of ...

### Cubic function - Wikipedia

determines what type of critical points the function has. If Δ 0 ... place as does the cubic curve; ... can be found by using the cubic function whose roots are ...

### Recent questions tagged cubic-function-with-graph - Mathskey.com

how do I draw cubic function as a dashed curve using the points (-4,3),(0,0),(2-6), and (5,5) ... cubic-function-with-graph;

### javascript - svg draw dashed bell curve between 2 points ...

svg draw dashed bell curve between 2 points. ... x2, and height and it should draw a dashed bell curve ... C \${quart} \${height}, \${quart} 0, \${quart*2} 0, \${quart*3 ...

### javascript - Dashed Curves on Html5 Canvas Bezier - Stack ...

... which requires knowing the arc length of your bezier curve at various points ... function(p0, p1) { return [p1[0 ... actually draw a dashed curve ...

### Plot expression or function - MATLAB fplot

This MATLAB function plots the curve defined by the ... use a line width of 2 points. For the second, specify a dashed red line style ... [0,1]; for example, [0.4 0.6 ...

### Cubic Function Graph - [email protected]

Cubic Function Graph. ... =\$a_3x^3 + a_2x^2+a_1x + a_0\$ is a cubic function with \$a_3\$ \$\neq\$ 0. ... Step 4: Draw a smooth curve joining all the points.

## Suggested Questions And Answer :

### use a graphical method to solve this simultaneous equation: y=x^2 and y=x+6

Plot y=x^2 and y=x+6 on the same graph. The first is a U-shaped curve sitting on the origin (0,0), the second is a straight line crossing the x axis (y=0) at x=-6 and crossing the y axis (x=0) at y=6. Where the line cuts through the U curve represents the solution to the equation x^2-x-6=0 (x^2=x+6). The solution to this quadratic is x=-2, +3. The intersection points are therefore (-2, 4) and (3, 9). The remaing part of your question involves straight lines only, linear equations. For (a) draw a straight line joining y=-4 on the y axis to x=2 on the x axis. Also draw a line joining y=3 1/3 to x=-1 2/3. These points on the axes are where x=0 and y=0 for the two functions. The lines don't stop at the axes, so just continue them after they cut the axes. What may have been confusing to you is that the first equation starts y=..., but the second equation has x and y together in an expression. However, you can move things around in an equation and, if you want, you can make the equation look like y=... or x=... or just combine x and y in an expression. It doesn't matter. What you'll find is that the lines are parallel (have the same slope) so that means they never cross and that means there's no solution. In (b) the two lines have different slopes so we would expect them to intersect. The first equation means joining (0,3) to (9/2,0) and (0,11/2) to (11/3,0). Again, extend the lines to beyond where they cut the axes. Note that the second equation simplifies to 3x+2y=11. The solution to the equations solved simultaneously or by substitution give a single intersection point at (3,1).

### graph the function y=3 sin 2 pi x

If you're used to thinking of degrees rather than radians, then remember that (pi) radians=180 degrees, so (pi)/4=45, (pi)/2=90, (pi)/6=30. And, since sin(0)=sin(180)=sin(360)=sin(540)=sin(720)=0, then sin(0)=sin((pi))=sin(2(pi))=sin(3(pi))=sin(4(pi))=0, and so on. This is the same as putting x=0, 1/2, 1, 3/2 in the function. These are the zeroes, and they repeat forwards and backwards (positive and negative) along the x axis at these regular intervals of 1/2. When we have sin((pi)/2)=sin(90)=1, x=(pi)/4(pi)=1/4. 3sin(2(pi)x) when x=1/4 is 3; sin(3(pi)/2)=-1, and x=3(pi)/4(pi)=3/4, so 3sin(2(pi)x) when x=3/4 is -3. The sine curve oscillates between 3 and -3 (amplitude) cutting the x axis at the above zeroes. To help draw the curve accurately it's useful to find out a few other points: for example sin(30)=sin((pi)/6)=1/2 and x=(pi)/12(pi)=1/12 and y=3/2 or 1.5; also y=3/2 when x=5/12 because sin(5(pi)/6)=sin((pi)/6)=1/2. When deciding on a scale for the axes you might consider subdivisions of 1/12 along the x axis between the units 0, 1, 2, 3, etc. these will make it easier to pinpoint 1/12 and 5/12 and other such values.

### I just need help getting started on this problem. I don't even know how to begin!

The graph is centered over the line x=-3. This line acts like a mirror so that the two halves of the inverted U-shaped graph are reflected in it. The graph has to be shifted 3 units to the right so that the line x=-3 becomes the line x=0, which is the y (f(x)) axis. The equation of the graph changes to f(x)=-0.5x^2+8. When x=0 f(0)=8, and this value on the vertical axis is the highest point of the curve (vertex). To picture f(x+k) you need a picture, a moving picture. Think of the horizontal axis, the x axis, and the line f(x)=8, which is a line parallel to the x axis a distance of 8 units above it. The two lines resemble a track, like a rail track. The curve  is, as we've established, an inverted U shape where the arms of the U are moving further apart the further away the curve is from its vertex. The curve has a constant shape and remains in contact with the line f(x)=8. We start with the curve right in the middle so that the vertical axis bisects the U curve. This is when k=0.  Now, we're going to slide the curve so that its vertex runs along the track f(x)=8 left and right. The curve cuts the axis at two points separated by 8 units, the gap between the zeroes of the function. When the curve is centrally positioned, the points on the x axis are -4 and 4 and k=0. Slide the curve to the left (negative side) and this is equivalent to positive values of k; slide it to the right and we're into negative values of k. If we move the y (f(x)) axis with the curve it becomes the movable axis of symmetry, the mirror I mentioned earlier. The points where the curve cuts the x axis remain 8 units apart. When the curve moves leftward one unit, the value of k increase by 1, and rightward one unit when k decreases by 1. The vertical line x=2 is unmovable, but as the curve slides from left to right this line touches or cuts through the curve. Consider only the part of the curve lying on or above the x axis. When does it touch the vertical line? Move the curve so that the right part of the curve just touches the line. This is the zero with the higher value. So it must be when x=2 is a root, i.e., -0.5(2+k)^2+8=0. 0.5(2+k)^2=8, so (2+k)^2=16, 2+k=+4, and k=4-2 or -4-2, which is 2 or -6. If we slide the curve past the line x=2, till the left part of the curve touches the line on the right side, the point where it touches is the other root, 8 lower than the the right-hand root. So if 2 was the right root then 2-8=-6 is the left root, and if 2 was the left root then 10 is the right root. So we have the functions f(x)=-0.5(x+2)^2+8 and f(x)=-0.5(x-6)^2+8, where the values of k have been substituted. When the curve is on the left of x=2, the roots of f(x+2) are -6 and 2 and when on the right the roots of f(x-6) are 2 and 10. In between these values of k, 2 and -6, the curve touches or is above the x axis, so this the range for k: -6 Read More: ...

### How are the locations of vertical asymptotes and holes different, and what role do limits play?

To talk about asymptotes and holes, you need pictures. These pictures are graphs of functions. The simplest function containing vertical and horizontal asymptotes is y=1/x, where x is the horizontal axis and y the vertical axis. The vertical asymptote is in fact the y axis, because the graph has no values that would quite plot onto the y axis, although the curve for 1/x gets very, very close. The reason is that the y axis represents x=0, and you can't evaluate 1/x when x=0. You' d have to extend the y axis to infinity both positively and negatively. You can see this if you put a small positive or negative value for x into the function. If x=1/100 or 0.01, y becomes 100. If x=-1/100 or -0.01, y becomes -100. If the magnitude of x decreases further y increases further. That's the vertical asymptote. It represents the inachievable. What about the horizontal asymptote? The same graph has a horizontal asymptote. As x gets larger and larger in magnitude, positively or negatively, the fraction 1/x gets smaller and smaller. This means that the curve gets closer and closer to the x axis, but can never quite touch it. So, like the y axis, the axis extends to infinity at both ends. What does the graph look like? Take two pieces of thick wire that can be bent. The graph comes in two pieces. Bend each piece of wire into a right angle like an L. Because the wire is thick it won't bend into a sharp right angle but will form a curved angle. Bend the arms of the L out a bit more so that they diverge a little. Your two pieces of wire represent the curve(s) of the function. The two axes divide your paper into four squares. Put one wire into the top right square and the other into the bottom left and you get a picture of the graph, but make sure neither piece of wire actually touches either axis, because both axes are asymptotes. The horizontal axis represents the value of x needed to make y zero, the inverse function x=1/y. Hence the symmetry of the graph. Any function in which an expression involving a variable is in the denominator of a fraction potentially generates a vertical asymptote if that expression can ever be zero. If the same expression can become very large for large magnitude values of the variable, potentially we would have horizontal asymptotes. I use the word "potentially" because there's also the possibility of holes under special circumstances. Asymptotes and holes are both no-go zones, but holes represent singularities and they're different from asymptotes. Take the function y=x/x. It's a very trivial example but it should illustrate what a hole is. Like 1/x, we can't evaluate when x=0. However, you might think you can just say y=1 for all values of x, since x divides into x, cancelling out the fraction. That's a horizontal line passing through the y axis at y=1. Yes, it is such a line except where x=0. We mustn't forget the original function x/x. So where the line crosses the y axis there's a hole, a very tiny hole with no dimensions, a singularity. So a hole can occur when the numerator and denominator contain a common factor. If this common factor can be zero for a particular value of x, then a hole is inevitable. Effectively it's an example of the graphical result of dividing zero by zero. With functions we can't simply cancel common factors as we normally do in arithmetic. Asymptotes and holes are examples of limits. Asymptotes can show where functions converge to a particular value without ever reaching it. Asymptotes can be slanted, they don't have to be horizontal or vertical, and they can be displaced from both axes. Graphs can aid in the solution of mathematical and physics problems and can reveal where limitations and limits exist for complicated and complex functions. Knowing where the limits are by inspection of functions also aids in drawing the graph. This helps in problems where the student may be asked to draw a graph to show the key features without plotting it formally.

### Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins

### How do you graph xy^2 + yx^2 + y-x=1 in calculus

First, let's put in a few values of x and y: x=0: y=-1; y=0, x=1; x=2, 4y+2y^2+2-y-1=0, 2y^2+3y+1=0=(2y+1)(y+1), y=-1/2 and -1; x=-1: -y^2+y-1-y-1=0, -y^2-2=0, no real value for y. This gives us the points (0,-1), (1,0), (2,-1/2), (2,-1). Note that there are two values of x when y=-1, two values of y when x=2. Treating the function as a quadratic in y: xy^2+y(x^2-1)+x-1=0, using the formula: y=(1-x^2+sqrt((1-x^2)^2-4x(x-1)))/(2x)=(1-x^2+sqrt(x^4-6x^2+4x+1))/(2x). When x=0 y=-1, but this formula otherwise covers all x<>0. It is a very unusual graph. Note that y is defined only when the square root is positive, that is, when the quartic is positive: x^4-6x^2+4x+1>0. This factorises: (x-1)(x^3+x^2-5x-1)>0. Differentiating the function: 2xy+x^2dy/dx+2xydy/dx+y^2+1-dy/dx=0, so dy/dx=-(y^2+2xy+1)/(x^2+2xy-1). So we can calculate dy/dx at each point: (0,-1): -2/-1=2 (1,0): -1/0  (2,-1/2): -(1/4-2+1)/(4-2-1)=3/4 (2,-1): -(1-4+1)/(4-4-1)=2/-1=-2 These are the slopes of the tangents at each of the 4 points. (1,0) is indeterminate suggesting a vertical tangent, which implies a vertex as in a parabola lying on its side. The slopes help to give some idea of the shape of the curve at various points. Notice where a negative slope is near to a positive slope, because there will be a turning point in between. dy/dx=0 at a turning point, and the condition for this is y^2+2xy+1=0, or x=-(y^2+1)/2y, bearing in mind that some values of x have no real y value. Also, y=(-2x+sqrt(4x^2-4))/2=-x+sqrt(x^2-1). The points above can be plotted with their tangents.

### For the function ∫_0^2▒〖f(x)=(x-2)^2+2 〗

For the function ∫_0^2▒  f(x)=(x-2)^2+2  dx we are going to estimate the area under the curve using the Simpson’s  rule where n = 6. The domain is [a, b] – [0, 2]. We have n = 6  intervals, with n+1 = 7 points with 7 f-values. (x0, f(x0), (x1, f(x1)), (x2, f(x2)), (x3, f(x3)), (x4, f(x4)), (x5, f(x5)), (x6, f(x6)). Δ = (b – a) = (2 – 0) = 2 h = Δ/n = Δ/6 = 2/6 = 1/3.  (i.e. there are n=6 intervals of length = 1/3) This gives us, x0 = 0, x1 = 1/3, x2 = 2/3, x3 = 1, x4 = 4/3, x5 = 5/3, x6 = 2. Using f(x) = (x – 2)^2 – 2, f(x0) = 2, f(x1) = 7/9, f(x2) = -2/9, f(x3) = -1, f(x4) = -14/9, f(x5) = -17/9, f(x6) = -2, Simpson’s rule is, A ≈ (h/3)(f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) + f(x6)) Substituting for h and the various values of f(x), A ≈ (1/9)(2 + 4(7/9) + 2(-2/9) + 4(-1) + 2(-14/9) + 4(-17/9) + (-2)) A ≈ (1/81)(18 + 4(7) + 2(-2) + 4(-9) + 2(-14) + 4(-17) + (-18)) A ≈ (1/81)(18 + 28 – 4 – 36 – 28 – 68 – 18) A ≈ (1/81)(-108) A ≈ -4/3

### how do you calculate the demand curve of (Q 40,000-500P)

This demand curve is a straight line. To draw it you normally make the price axis P vertical and the quantity Q axis horizontal. To draw the line work out the price when the quantity is zero (no demand). This happens when Q=0 and 500P=40000, so divide both sides by 100 by removing two zeroes: 5P=400 then divide through by 5: P=80. So on the P axis mark 80. Although it wouldn't be realistic to have a zero price, nevertheless it will enable you to draw the line. When P=0 Q=40000, so mark a point on the Q axis to indicate 40000. You can calibrate both axes to fit the values, so the Q axis could be measured in units of 1000: 0, 1000, 2000, and so on. Or you could use units of 5000: 0, 5000, 10000, 15000, ..., 40000. and so on. Join these points on the two axis to produce your demand line. You can then read off Q and P values on the line. For example, at Q=20000, P=40.