How do you form a parabola from a straight line?
458048
Or you can do this:\n\nhttp://www.xahlee.org/SpecialPlaneCurves_dir/Parabola_dir/parabolaTangents.png\n\nto form a parabola out of many straight lines.
301287
I am guessing you mean the area between the line y = 1-2x and the parabola y = x^2 + 2x + 1.\n\nTo do this, it takes a little calculus.\n\nFirst you find the points of intersection between the parabola and the line by setting the equations equal to each other.\n\nx^2 + 2x + 1 = 1 - 2x\n\nWe get x^2 = 4x = 0\nor x(x + 4) = 0 so the points of intersection occur at x = 0 and x = 4.\n\nAt this point you need to integrate each equation and evaluate between -4 and 0.  Subtract the area of the parabola between -4 and 0 from the line between -4 and 0 and you get the answer.\n\nThe integration for the line gives x - x^2 between -4 and 0\nThe integration for the parabola gives 1/3x^3 + x^2 + x between 0 and -4\n\nThis results in 20 - (64/3 - 16 + 4)= 32/3 square units\n\nCheck my math, but I think it's okay
384702
The first 3 are all straight lines, so you just need to get 2 points in each.\n1. y = |x| + 3 for x = 1 then y = 4, for x = 2 y = 5.\n\n2. y = 3/2x - 2 for x = 2 then y = 1, for x = 4 y = 4\n\n3. y = x - 5 for x = 5 y = 0 for x = 6 y = 1\n\nFor #4. This is a curve shaped like and upside down U that passes through 0,0, for all values of x, y will always be negaive.\n\nThe best way is to substitute values of x and get values of y and sketch.\n\nHope I was some help
497708
to find if a line is paralel to each other, you had to find out it's gradient or slope. If the gradient is the same, then the lines are paralel.\n\nTo find gradient is rather easy.\n\nYou just plug in the points into (y2 - y1)/(x2 - x1)\n\nSo the first line is:\n(3 - - 7) / (1 - - 4)\n= (3 + 7) / (1 + 4)\n= 10/5\n= 2\n\nand the second line is:\n(10 - 6) / (4 - 2)\n= 4 / 2\n= 2\n\nyou see here that 2 = 2, so the two lines are paralel
831482
We just started this stuff in school.  first, think of f(x) as y.  then make a table with the left column as x, and the right column as y.  the vertex is going to be -b/2a.  so in this case.  it'll be -1/4.  that's the middle number in the table and also the vertex.  now put above that number in the table on the x column as so.\nx  y\n1\n0\n-1/4\n-2.5\n-3.5\nnow fill in the the x column by plugging in the x variables.  plug in 0 for x and thats your y-intercept.  then plug in your 0 for y, and thats your x-intercept.  this is going to form a parabola so the axis of symmetry is -1/4.  the domain is going to be all real numbers because there are no contradictory numbers to fill in.  Im not sure how to do the range because im only a junior in high school but I hope I helped a little.
716583
If you know the vertex then you know\n\na (x-xv)^2=y-yv  , use P to find a, the shape \n\nat P:\na(xp-3)^2 =yp-1\na(-1-3)^2 = 3-1\n16a = 2\na=1/8\n\nSo, the parabola is \n\ny = 1/8*(x-3)^2 + 1\n\nCheck: \nat V  y= 1 ok\nat P y = 1/8(-4)^2 + 1 = 16/8+1 = 3 ok
374950
Plug in x, solve for y as follows:\nIf x=0 then solve for y and get y= 3/2\nif x=10 y= -10/2+ 3/2 = -7/2\nCreate a graph using the horizontal as the x scale and the vertical as y.  Plot the above and you will see that it's a straight line.
291224
The simplist equation that will give a parabola is\n\nY = X squared (ie X times x) don't know how to typy that in .\n\nOn the number plain the lowest value for this equation is (0,0). this is obtained by substituting '0' in the equation for X. There is no maximum value, unless you consider 'infinity' a maximum value. This is because you can keep increasing the value for X untill you reach infinity, and infinity+1 doesn't exist\n\nA parabola is also the locus of a point that moves so that is always equidistant fron a straight line and a point. In the case of Y=X^2 The point is (0,-1) and the straight line is Y=-1
184282
y = 7\n\na horizontal line
942923
Without using calculus, draw the rectangle with a pretty river for one side.  Label the two congruent ends of fence x.  Since they add up to 2x, the remaining piece of fence is 5000 - 2x long.\n\nThe area is now x(5000-2x) or 5000x - 2x^2\n\nIf you graphed this it would be an upside down parabola,and its highest point would be the vertex.  The formula for the x coordinate for the vertex is x = -b/2a\n\nFor this, it's x = -5000 over 2(-2) or -5000/-4 or 1250\n\nSo x = 1250 and the two ends are 1250 ft each leaving 2500 ft for the third side.  \n\nHope that helped.
972913
It's actually all explained for you right there in your "question". Maybe it would help you to draw the line on a piece of graph paper. Plot the point (3,5), then the point (-2,0) and draw a straight line through them. It tells you that the slope (m) is 1. So for every step to the right, the line goes up one step. The point (3,5) is the same as (x1,y1) and (-2,0) is  the same as (x2,y2). So you replace 3 for x1 in the point-slope form equation, and so on. It really helps to pay attention on class.
823299
How about drawing a bunch of straight lines of differing length, from a fixed point, so that the tips create the image of an arc?
901295
for the first part...if a is a negative value then the parabola will be drawn downwards (somewhat like a sad face) and likewise if the a value is positive the parabola will be drawn upwards like a smiley face.\n\nto find the y intercept substitute x=0 into the equation...so the y intercept is equal to a multiplied by the square of negative h plus k.\n\nthe turning point is represented by (h,k) the x value of the turning point is equal to h and the y value is equal to k.\n\nhope that helped.
715400
Angle of line from horizontal position.
760976
The parabola y = (x+3)(x-5) crosses the x axis at -3 and 5.  It opens upward, so its vertex must be below the x axis.  So y < 0 for x in (-3, 5).
358399
There is an equation for this, but I don't remember what it is.  Guess and check with a calculator is funner anyways...\nI didn't show much work (how much work is there with guess and check), but I think these are correct. (&=infinity)\n\n1. The range of k is [-2,&)\n\n2. When k=3 then the ratio of a:b will be 1:5.  The equation of the parabola will be y=-X^2 + 4X + 5.\n\n3. I wasn't sure exactly what you were asking on Q 3.  If you are looking for where the tangent line of the parabola at x=0 crosses the x-axis, the point will be (-1.25,0).  If you are saying that from (A,0) to (B,0) in Q 2 is the diameter of a circle, and you are looking for where the tangent line of that circle at x=0 intersects the x-axis then the point is (-4.5,0).\n\nI hope that helps.
1112480
Carefully, very carefully :)\n\nf(x) =-1/3x^2 is a parabola origin at (0,0) with branches downward.\n\n\nf(x)=x^2-6x+9 =(x-3)^2 orgin at x=3, y intecept at x=0