Sunday

April 20, 2014

April 20, 2014

Number of results: 122

**math**

weloo Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the area enclosed by the two parabolas. (c) Use Mathcad to draw the...
*Tuesday, May 8, 2012 at 6:11pm by weloo_volley*

**math**

hilp meeee trying .... Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the area enclosed by the two parabolas. (c) Use ...
*Tuesday, April 24, 2012 at 11:53pm by weloo_volley*

**math**

hilp meeee trying .... Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. (a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points. (b) Compute the area enclosed by the two parabolas. (c) Use ...
*Monday, April 30, 2012 at 3:56am by weloo_volley*

**Math**

Hi Anna. Didn't have the time to actually answer the question that you wanted answering but i thought i'd direct you to a very good website that would give you a re-cap about Parabolas if that is ok. It is: Parabolas Lessons 1 to 4 by Jack Sarfaty. You should find Jack Sarfaty...
*Monday, September 22, 2008 at 6:25pm by STEVE (from England)*

**Pre-Calc**

We could write our answer in the form y+2 = 5(x-4)^2 take a look at http://jwilson.coe.uga.edu/emt725/class/sarfaty/emt669/instructionalunit/parabolas/parabolas.html half-way down the page at example 3 and follow the steps.
*Wednesday, May 2, 2012 at 10:17pm by Reiny*

**Algebra 2**

I am creating a project for math in which I have to draw an object using: parabolas, hyperbolas, circles,etc. We use conic section equations to identify where they are on the graph. I have confusion about writing the equation of parabolas. I know the equation is 4px=y^2 but I ...
*Tuesday, June 5, 2007 at 9:46pm by mariya*

**math**

Question-2; Consider the function f(x)= -cos3x -4sin3x. (a)Find the equation of the line normal to the graph of f(x) when x= pie/6 . (b)Find the x coordinates of the points on the graph of f(x) where the tangent to the graph is horizontal. (c)Find the absolute extrema of the ...
*Saturday, April 21, 2012 at 6:36pm by weloo_volley*

**Math**

Find the points of intersection of the parabolas y=1/2x^2 and y=1-1/2x^2. Show that at each of these points the tangent lines to the two parabolas are similar Answer (+-1,1/2) Please show and explain all steps thank yo
*Wednesday, November 6, 2013 at 7:02pm by Jason*

**Parabolas in standard form!**

Hello, I am having the worst time trying to solve these parabolas and putting them into x-h = a(y-k)^2 form. :( There are two problems that i keep doing something wrong. could someone solve them so I can have a set up for the rest of my problems? :) 3y^2+6y+108x-969=0 and x^2+...
*Thursday, May 24, 2012 at 9:06pm by Kate*

**calculus**

After making a diagram, I cannot tell from the question if you mean the smaller part in quadrant II or the larger part of the region between the two parabolas. the two parabolas intersect at (2,4) and (-2,4) and the line intersects the y=x^2 at (-2,4) and (6,36) and is tangent...
*Wednesday, December 1, 2010 at 6:38pm by Reiny*

**Algebra-Parbola**

Factoring doesn't help much. An easy method of graphing parabolas is to make a T-chart (with x and y). Pick a few x-values (like -2, -1, 0, 1, and 2) and then calculate the y-values. I wouldn't really recommend this. Learning the rules for graphing parabolas will be much more ...
*Friday, November 23, 2007 at 9:32pm by Michael*

**Math**

1. What is the vertex of each of these parabolas? a. y = (x –7)2 +4 b. y = 3(x + 7)2 – 4
*Wednesday, November 13, 2013 at 9:55pm by Sammy*

**algebra 2**

what is a parabolas, circles, ellipse, or hyperbolas
*Tuesday, December 15, 2009 at 4:43pm by breanna*

**Pre calc**

Projectiles are always parabolas
*Monday, May 6, 2013 at 9:07pm by Anon*

**calculus **

find the points of intersection of the parabolas Y=(1r2 x^2) and Y=10x - 2.
*Thursday, February 18, 2010 at 6:31pm by Anonymous*

**calculus **

find the points of intersection of the parabolas Y=(1r2 x^2) and Y=10x - 2.
*Thursday, February 18, 2010 at 6:50pm by Anonymous*

**Algebra: Parabolas**

So it doesn't matter what point is given as long as it's on the parabola?
*Wednesday, April 25, 2012 at 10:04pm by Anonymous*

**parabolas**

Find the equation and sketch the parabola that has vertex V (-1; 0) and focus F(-4; 0).
*Monday, May 6, 2013 at 9:59pm by dylan*

**math**

there are ___ parabolas that have a vertex (-3,-4) and pass through the point (-1,4)
*Monday, May 23, 2011 at 7:52pm by Anup*

**Math **

Give three examples of the equations of parabolas that have (-2,0) and (6,0) as x-intercepts.
*Friday, March 2, 2012 at 12:22am by Nicole*

**calculus**

the parabolas intersect at x = -2,2 area is thus ∫[-2,2] (4-x^2) - (x^2-4) dx = 64/3
*Thursday, October 3, 2013 at 6:40pm by Steve*

**MATH HELP**

write the equations of two different parabolas whose vertices are at (3,2).
*Wednesday, November 13, 2013 at 9:56pm by vanessa*

**Parabolas in standard form!**

WOW! Thank you so much! you made more sense than my teacher! :D
*Thursday, May 24, 2012 at 9:06pm by Kate*

**Calculus Help!!!**

Find c > 0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 13.
*Thursday, November 21, 2013 at 10:32pm by Ryan*

**Calc**

I made a sketch, which showed the 2 parabolas don't intersect. so you have the open space between them cut off by the x-axis (y=0) and the horizontal y = 4 A vertical form (2,0), on the second and (2,4), which lies on the first, splits the area into two equal regions. There is...
*Thursday, April 2, 2009 at 1:10am by Reiny*

**Precal**

What is the standard form of this parabola: -2x^2+16x+24y-224=0 Please Help, parabolas are evil.
*Friday, May 25, 2012 at 3:36am by Sam*

**Math**

If f(x)=x^2-6x+14 and g(x)=-x^2-20x-k, determine the value of k so that there is exactly one point of intersection between the two parabolas.
*Tuesday, August 17, 2010 at 4:58pm by Emily*

**Math**

If f(x)=x^2-6x+14 and g(x)=-x^2-20x-k, determine the value of k so that there is exactly one point of intersection between the two parabolas.
*Thursday, October 21, 2010 at 4:55pm by Emily*

**Precalculus**

Having had algebra I, you already know all about parabolas. The vertex is where it changes direction. Use that knowledge.
*Tuesday, June 25, 2013 at 11:13am by Steve*

**Calculus Help**

Let c and dd be positive real numbers. Show that regardless of c,d, the parabolas r=c/(1+cos(theta)) and r=d/{(1-costheta) intersect at rights angles
*Friday, December 3, 2010 at 12:00am by Sarah*

**algebra (parabolas)**

find the axis of symmetry and vertex of the function. tell whether it goes upward or downward. y=3*xsquared+6x-2
*Sunday, March 20, 2011 at 6:27pm by Bri*

**algebra1 HELP PLEASE**

by now you know about parabolas. h(t) = 160t - 16t^2 the vertex of a parabola is at t = -b/2a, which here is t = 160/32 = 5 So, at t=5, h(t) = 400
*Monday, July 8, 2013 at 11:34am by Steve*

**math**

We are learning about parabolas and the question was using partial facotring to find 2 points that lie on the parabola. How would I do that? Ex. y=x^2-12x+23
*Thursday, January 24, 2008 at 7:43am by Lena*

**algebra (parabolas)**

find the axis of symmetry and vertex of the function. tell whether it goes upward or downward. 1. y=-xsquared+9 2. y=-2xsquared=7x-21
*Sunday, March 20, 2011 at 7:06pm by Bri*

**Algebra: Parabolas**

Since it's a standard y^2 = ax, then we know that 16 = 3a a = 16/3 which makes p = a/4 = 4/3
*Wednesday, April 25, 2012 at 10:04pm by Steve*

**Calculus**

It's a hyperbolic paraboloid for y=k, x = k^2-4z^2 for z=k, x = y^2-4k^2 both are parabolas for x=k, y^2-4z^2 = k^2 is a hyperbola
*Wednesday, May 15, 2013 at 1:33pm by Steve*

**Calc 2**

Find c>0 such that the area of the region enclosed by the parabolas y=x^2−c^2 and y=c^2−x^2 is 90.
*Friday, February 10, 2012 at 7:52am by Rebecca*

**Calculus 2**

Find c>0 such that the area of the region enclosed by the parabolas y=x^2−c^2 and y=c^2−x^2 is 90.
*Friday, February 10, 2012 at 2:15am by Rebecca*

**Math/Algebra**

Does anyone know how to graph quadratic functions (with parabolas). Could someone please help me!!Could you help me with these two problems don't forget to make a chart for x, the equation, the y, and the coordinates. 1. Y=2x^2 2. Y=-5x^2
*Monday, March 30, 2009 at 4:27pm by Smores*

**algebra,math**

Tracy, You have been asking some very basic and simple questions about parabolas. If you are studying these right now, you should know these answers.
*Monday, October 24, 2011 at 11:02pm by Reiny*

**Calculus**

A rectangle is insribed between the parabolas y=4(x^2) and y=30-x^2. What is the maximum area of the rectangle? Domain: [-3,3] Range:[-2, 40] THANK YOU!
*Tuesday, December 7, 2010 at 10:34pm by Anna*

**calculus area between three curves**

also help with Find c>0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 430.
*Monday, December 2, 2013 at 2:12pm by jaime*

**Graphs&Intercepts**

Hard to say. Linear equations, parabolas, hyperbolas, what? There are lots of different functions, which have very little in common regarding the effect of negative x-coefficient.
*Thursday, October 13, 2011 at 5:08pm by Steve*

**algebra**

what is the standard equation of the following parabolas?! 1. passes through (-5,1) 2. Direction 2x-9=0 3.E(4,2) as end of LR 4. Focus at (3,0) 5. LR=10 pls answer
*Tuesday, April 27, 2010 at 1:50am by Anonymous*

**Parabolas in standard form!**

108x = -3y^2 - 6y + 969 108x = -3(y^2 + 2y + 1) + 972 x = -1/36 (y+1)^2 + 9 I'm not too sure how to do the second one. I'm really sorry.
*Thursday, May 24, 2012 at 9:06pm by Neil*

**Pre-Calculus**

I'm currently working with parabolas, and I'm not sure if I am doing this specific problem correctly.. 12. x= y^2-6y+6 y= -b/2a = 6/2(1)= 6/2= 3 x= (3)^2+6(3)+6 x= 33 y=3 (33,3) <--- what I graphed
*Friday, May 10, 2013 at 4:20am by Caroline*

**Calculus kinda with the area of two curves**

II don't even know where to start with this can anyone help?!? Find c>0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 270.
*Monday, December 2, 2013 at 2:42pm by Roni*

**AP Calculus**

A rectangle is inscribed between the parabolas y=4x^2 and y=30-x^2. what is the maximum area of such a rectangle? a)20root2 b)40 c)30root2 d)50 e)40root2
*Tuesday, December 14, 2010 at 10:19pm by Jim*

**Parabolas in standard form!**

Can you solve this one?: x^2-12x-48y-372=0 & I made a mistake one the second one!! Its actually: x^2+14x+44y+313=0 !!
*Thursday, May 24, 2012 at 9:06pm by Kate*

**Algebra**

Both the right side up and the upside down parabolas defined by y = x^2-5x+6 and y = -(x^2-5x+6) saisfy the requirements [(x-6)(x+1)]^2 = 0 x=6 twice x=-1 twice
*Tuesday, July 7, 2009 at 12:27am by Damon*

**parabolas**

The distance between the vertex and the focus is 3 so the directrix must be x =2 (from -1 to 2 is 3) let (x,y) be any point on our parabola √( (x+4)^2 + y^2) = √( (x-2)^2 + 0) square both sides and expand x^2 + 8x + 16 + y^2 = x^2 - 4x + 4 y^2 = -12x - 12
*Monday, May 6, 2013 at 9:59pm by Reiny*

**Algebra**

generally speaking, linear functions yield graphs that are straight lines, quadratic functions yield parabolas, .... If you give me a specific example ...
*Tuesday, June 10, 2008 at 8:34pm by Reiny*

**algebra (parabolas)**

tell whether the graph opens upward or downward. then find the axis of symmetry and vertex of the graph of the function 1. y=xsquared-5 2. y=-2*xsquared+6x+7
*Sunday, March 20, 2011 at 6:18pm by Bri*

**algebra (parabolas)**

3 x^2 + 6 x = y + 2 x^2 + 2 x = (1/3) y + 2/3 (2/2)^2 = 1 so add 1 (which is 3/3)to both sides x^2 + 2 x + 1 = (1/3)y + 5/3 (x+1)^2 = (1/3)(y+5) symmetric about the line x = -1 vertex at (-1,-5) when x gets big, y gets big, opens up (holds water)
*Sunday, March 20, 2011 at 6:27pm by Damon*

**Grade 12 Calculus**

you don't need calculus for this. It can just confirm your answers. these are just ordinary parabolas. Find the vertex, and check the values at the ends of the intervals.
*Tuesday, April 16, 2013 at 12:15pm by Steve*

**math**

region bounded by the parabolas y=x^2 and y=6x-(x^2) is rotated about the x-axis so that a vertical line segment cut off by the curves generates a ring. find the value of x for which we obtain the ring of largest area
*Monday, January 28, 2008 at 3:01pm by ML*

**Trig**

A parabola has its vertex on the graph of the line y=3x+1 and passes through (1,10). If it is the same size,? shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas Thanks!!!
*Wednesday, April 24, 2013 at 6:39am by Rebekah*

**Pre calc**

A parabola has its vertex on the graph of the line y=3x+1 and passes through (1,10). If it is the same size,? shape, and direction as the graph of y=3x^2, find the equation (s) of all possible parabolas
*Tuesday, April 23, 2013 at 8:35pm by Rebekah*

**Algebra **

Graph parabolas how would i do this y=x^2 + 1 (-3 < or equal to X < or equal to 3)
*Monday, April 2, 2012 at 7:04pm by BIBIN*

**multivariable calculus**

2x^2 - 3z^2 - 4x - 4y - 12z + 2 = 0 2(x-1)^2-2 - 3(z+2)^2+12 - 4y + 2 = 0 2(x-1)^2 - 3(x+2)^2 - 4y + 12 = 0 4(y-3) = 2(x-1)^2 - 3(z+2)^2 Looks like an hyperbolic paraboloid with vertex at (1,3,-2) The traces are an hyperbola in the xz plane, and parabolas in the others.
*Tuesday, March 26, 2013 at 8:33am by Steve*

**Algebra: Parabolas**

That's not quite true. They gave the added information that it's a standard parabola with the x-axis as its axis of symmetry, and (0,0) as its vertex. In general, knowing only one point on the curve is not enough, but with what they said, you know that you also have (0,0) and...
*Wednesday, April 25, 2012 at 10:04pm by Steve*

**Calculus and vectors**

of course. All parabolas have the vertex midway between the roots. The quadratic formula makes this easy to see, since the roots are at x = -b/2a ± (some junk) Starting from a nonzero height just places the smaller root at negative value.
*Monday, December 16, 2013 at 8:53am by Steve*

**Calculus (Continuity and Differentiability)**

In that case, the derivative is continuous. The 2nd derivative will not be continuous at the cusp. The function whose derivative is the v-shaped graph, will also be continuous, but display a sharp corner where the two parabolas intersect.
*Saturday, November 12, 2011 at 3:43pm by Steve*

**Calculus**

Start with sketching the solid/cross section. Cylinders are easier to sketch. Draw the trace (projection on the x-y plane. The shape stays the same for the full height. The trace has the two parabolas meet at (-1,0) and (1,0). Draw the two planes. Start with x+y+z=2 x=-1,y=0...
*Thursday, February 16, 2012 at 5:16pm by MathMate*

**math**

Not even having graphed this, I can tell that the second graph is simply the first one moved upwards 4 units Try it, it is really simple. Just find some ordered pairs for each and join them with a smooth line. (You should get two parabolas, both opening downwards, do one at a ...
*Monday, April 8, 2013 at 9:58am by Reiny*

**math**

Graph of Y = x^2. (X , Y) (-2 , 4) (-1 , 1) V(0 , 0) (1 , 1) (2 , 4). Add -4 to each value of X. Add 3/2(1.5) to each Y value: (X , Y) (-6 , 5.5) (-5 , 2.5) V(-4 , 1.5) (-3 , 2.5) (-2 , 5.5) Graph the two parabolas on the same graph for better comparison. V(-4 , 1.5), P(-3 , 2...
*Thursday, December 9, 2010 at 7:16pm by Henry*

**calculus**

They are both parabolas that "open up" in the +y direction, but the vertices are in different places. y = x^2 has a vertex (minimum) at (0,0), and the other function has a vertex at (-10,-2) The y = x^2 function opens up more slowly, which means it is narrower at a given ...
*Saturday, May 22, 2010 at 12:33am by drwls*

**Algebra 1 Question**

by this time you may know that parabolas have an axis of summetry. x-values equally distant from that axis produce the same y values. For example, you know that 2x^2-7 is symmetric about the y-axis, because x^2 is always positive, whether x is positive or negative. So, if you ...
*Wednesday, October 30, 2013 at 5:28pm by Steve*

**Calculus**

express C as a function of x and y: C = ye^-x find y' implicitly y'e^-x - ye^-x = 0 y' = y so, for any point on the curve Ce^x, the slope is just y On a perpendicular trajectory, the slope is -1/y dy/dx = -1/y y*dy = -dx y^2/2 = -x+c Thus parabolas opening to the left are ...
*Monday, March 5, 2012 at 10:43am by Steve*

**Math**

excuse me? You're in calculus, and don't know what a parabola is? Regardless, just work with the math. y'(x) = 6x-5 y'(2) = 7 y(2) = 2 So, you want the equation of the line passing through (2,2) with slope 7: (y-2) = 7(x-2) Your answer is correct. Any quadratic equation ...
*Sunday, October 28, 2012 at 10:55am by Steve*

**Math**

Rolle's Theorem states that on the interval (2,3) there will be a point c where f'(c) = the slope of the chord from f(2) to f(3) So, f(2) = 102 f(3) = 102 so, the chord has slope 0. so, at some point c in (2,3) f'(c) = 0 f'(t) = 80 - 32t f'(t) = where t = 80/32 = 2.5 That's ...
*Thursday, March 8, 2012 at 3:21pm by Steve*

**Integrated Math 1**

The parabolas y = -3x2 + 10x - 6 and y = -3x2 - 17x + 2 intersect at _____? A.the line y = 27x - 8. B. x=the square root of 2 over 3 and xthe -square root of 2 over 3. C. x=8/27. D. they do not intersect.
*Sunday, July 8, 2012 at 1:13pm by Tifini*

**Calc**

Even though they don't intersect, there is the area bounded between y=0 and y=2 Did you notice that the two parabolas are congruent and the second is merely translated 2 units to the right? so the horizontal distance between corresponding points is always 2 that is x2-x1 = (y-...
*Monday, October 10, 2011 at 2:27am by Reiny*

**College algebra**

A fountain in a shopping mall has two parabolic arcs of water intersecting. The equation of one parabola is y = -0.25x^2+2x and the equation of the second parabola is y=-0.205x^2+4x-11.75. How high above the base of the fountain do the parabolas intersect? All dimensions are ...
*Saturday, May 5, 2012 at 1:47am by Missy*

**Math: Polynomial inequalities**

Inequalities involving parabolas can get interesting. Solve as an equation: 2t^2 - t -3 = 0 (t+1)(2t-3) = 0 t = -1 or 3/2 Think now of the graphs of 2t^2 - 3 and t You want the region where the parabola is below the line. That is, -1 <= x <= 3/2 Of course, you can do ...
*Tuesday, October 18, 2011 at 3:34pm by Steve*

**Functions**

State the domain and range of each function. Justify your answer a) y=1/(x+5) b) y=√1-2x I know how to do domain and range, but I don't remember what shapes these form. It's not parabolas, is it? Then, to justify them, I need to draw them, so that's why I need to know ...
*Tuesday, September 8, 2009 at 8:33pm by Sally*

**Pre Calc**

It will be a parabola. The vertical axis will be at x = -3. F is the focal point. The vertex of the parabola is equidistant from F and the y = -2 line, at x = -3, y = 1. y = a (x+3)^2 + 1 is the equation. Do some reading up on parabolas to figure out what a is. I think you ...
*Tuesday, September 14, 2010 at 12:18am by drwls*

**Math**

surely your text has many examples of quadratic functions and their graphs. I don't know what you mean by "work" this problem. To graph the function, pick a few values of x, calculate the corresponding values of f(x), and plot the points. To get you started, f(x) = (x+3)^2 - 4...
*Tuesday, June 19, 2012 at 5:54pm by Steve*

**Alg 2 (ASAP)**

I have a parabola vertex (11,3) p is not given, focus pt is not given. I need to find the equation of the parabola. How would I do this w/o p value or focus pt? you cant. ̃here are an infinite number of parabolas that have that vertex.
*Tuesday, June 5, 2007 at 10:29pm by masha*

**algebra (parabolas)**

to find the axis of symmetry you must use the formula x=-b/2a which in (1. a=1 b=0 c=9) 1.x=-b/2a = o/1(2) 0/2 =0 your axis of symmetry would be 0 and to find out if it goes upward or downward you must plug in more numbers to get more cordinates
*Sunday, March 20, 2011 at 7:06pm by Ana*

**algebra**

you can probably save some time by looking at the points. You know that parabolas are symmetric, so, since y(-2) = y(4), the axis of symmetry is at x=(-2+4)/2 = 1. So, with x=1 the axis of symmetry (and hence at the vertex), you will have y = a(x-1)^2+k at x=0, y= -4, so -4 = ...
*Monday, October 29, 2012 at 3:54pm by Steve*

**Calculus**

c does not equal -12: a(3)^2 + b(3) + c = 0 Solve: c = -9a - 3b (not -12) For the given parabola, since the vertex is (3,0), and knowing that parabolas are symmetrical with respect to the vertex, we therefore have another point (2,2), the mirror image of (4,2) about x=3 ...
*Tuesday, September 21, 2010 at 2:47pm by MathMate*

**Mathematics**

The two parabolas should be easy to graph to see what you are finding. You will first need to find their intersection, x^2 + 2 = 10 - x^2 2x^2 = 8 x = ± 2 The height of the region is (10-x^2) - (x^2+2) = 8 - 2x^2 Because of the symmetry we will find the area from 0 to 2 and ...
*Sunday, March 21, 2010 at 9:53am by Reiny*

**Algebra HELP!**

I expect that you mean y = x^2 and f(x) = (x-2)^2 + 3. We suggest you to use a ^ before all exponents to help us understand what you are writing. Both curves are parabolas that open upward. The first one, "y" has its minimum value when x = 0, and the y value there is y = 0. ...
*Friday, March 28, 2008 at 11:13am by drwls*

**pre-calculus**

the problem here is that after having solved the equation, you now know where the graph crosses the x-axis. You want to know where 16(x^2-1) > 16x x^2 - 1 > x x^2 - x - 1 > 0 Now from what you know about parabolas, it should clear that the graph is above the x-axis ...
*Saturday, September 22, 2012 at 4:24pm by Steve*

**Math**

i am working with completing the square in parabolas and there's this word problem i just cannot solve..a farmer wants to make a rectangular corral along the side of a large barn and has enough materials for 60m of fencing. Only 3 sides must be fenced, since the barn wall will...
*Tuesday, December 9, 2008 at 8:35pm by Callie*

**Algebra 2**

(y-5)^2 = 5 ( x + 8) y gets big + as x gets big + so opens to the right well, all that is left is a focus and directrix. parabolas do not have major and minor axes like ellipses. if in the form (y-k)^2 = 4 a (x-h) so here a = 5/4 then vertex at (h,k) vertex to focus = a so at ...
*Monday, May 19, 2008 at 8:11pm by Damon*

**Math**

I assume you mean the monkey leaped straight up. Ignoring the fact that free-fall trajectories are parabolas, and not straight lines, we have x = height of leap 300 = hypotenuse (because the distance down the tree and over to the well is 300) √(200^2 + (100+x)^2) = 300 x...
*Sunday, July 15, 2012 at 11:34pm by Steve*

**Algebra--Math matePlease help**

I have not seen the diagram, but when presenting parabolas in the form f(x) = a(x-h)²+k it is usual to consider the starting point with the vertex at (0,0) and as we vary h and k, it will be translated to the final position. Back to the question, is the vertex drawn at (-...
*Wednesday, September 22, 2010 at 9:21am by MathMate*

**calculus**

a solid lies between planes perpendicular to x-axis at x=0 and x=17,cross section perpendicular to axis interval 17 greater or equal x & x greater or equal 0 are squares of diagonals go from parabolas y=-2x^(0.5) & y=2x^(0.5). find the volume of the solid
*Tuesday, March 27, 2012 at 7:05am by Anonymous*

**parabolas**

how do i find the equaion of the following parabolas? 1)intercepts at (2,0),(8,0) and (0,48) 2)turning point at (1,-2) and passing through the origin. can someone please teach me how to do this??thanks! The first: Write you standard equation of a parabola (it has two x ...
*Sunday, May 20, 2007 at 8:49am by Emma*

**calculus**

Your function is a simple parabola All parabolas have one turning point, called the vertex because of the + understood in front of f(x) = +x^2 .... the parabola opens upwards so it rises to infinity in both the 1st and 2nd quadrants. If the function had been f(x) = -x^2...
*Tuesday, December 18, 2012 at 9:40pm by Reiny*

**math**

x^2 means x squared, or x². If you just started with parabolas, it won't hurt to say what you've learned so far. Let's start over: We first assume the vertex of the parabola is at x=0. We know that y=0 at x=±24, or y=a(x-24)(x+24). We also know that y=12 at x=0, i....
*Friday, August 31, 2012 at 10:41am by MathMate*

**Physics**

The equation for maximum height is H = (V^2/2g)* sin^2 A The equation for range is R = (V^2/g)*sin(2A) where A is the launch angle, measured from horizontal. Note that both are proportional to V^2. Deriving the formulas would be a useful exercise for you. Both trajectory ...
*Monday, May 24, 2010 at 11:12pm by drwls*

**Algebra: Parabolas**

I have a math problem that requires me to find the focal length of a parabola using the information given on a graph. It is a standard horizontal parabola, the formula being y^2 = ax. The focus point is formula a = 4p. I know that a is positive, because the parabola opens ...
*Wednesday, April 25, 2012 at 10:04pm by Anonymous*

**calculus 2**

The two parabolas enclose an area between (0,0) and (1,1) y = x^2 or x=sqrt y is on the right and lower there. y = x^(1/2) or x = y^2 is above and left there. (All in quadrant one, no worry about signs ) the radius to x = -3 is (3+x) so outer radius = 3 + sqrt y inner radius...
*Tuesday, February 16, 2010 at 7:36pm by Damon*

**Calculus**

A function f(x) is said to have a removable discontinuity at x=a if: 1. f is either not defined or not continuous at x=a. 2. f(a) could either be defined or redefined so that the new function IS continuous at x=a...
*Sunday, February 6, 2011 at 12:47am by Abigail*

**college algebra**

Not quite. You found (sort of) when the height is greatest (the vertex of the parabola). That was not the question. You have h=-9.8t^2+88.2t when is h=137.2? (Aside. You sure about that 9.8? h = vt - 1/2 at^2 and a = 9.8) 137.2 = -9.8t^2 + 88.2t t = 2,7 So, at those two times...
*Monday, February 25, 2013 at 12:55am by Steve*

**algebra (parabolas)**

x^2 = (y + 5) when x gets big, y gets big, opens up (holds water) symmetric about y axis (x = 0) vertex at (0,-5) 2 x^2 - 6 x = - y + 7 x^2 - 3 x = -(1/2)y + 7/2 add (3/2)^2 = 9/4 to both sides x^2 -3 x +9/4 = -(2/4)y + 23/4 (x-3/2)^2 = -(1/4)(2y-23) axis at x = 3/2 vertex at...
*Sunday, March 20, 2011 at 6:18pm by Damon*

**calculus**

we are looking at two intersecting parabolas, one opening up the other down find their intersection points: 2x^2 - 9 = 18 - x^2 . . x = ± 3 in the region we want the downwards opening curve is ABOVE the upwards curve, so the effective height in our region is 18-x^2 - (2x^2-9...
*Sunday, January 4, 2009 at 4:08pm by Reiny*