Post by: Cammmeron! on March 20, 2013, 09:07:48 pm
1.For the quadratic with rule y = 2x^2 + mx+ 4, find the values of m for which there is:
a.one solution
b.two solutions
2.Find the coordinates of the points of intersection of the line with equation y = x + k and the parabola with equation y = x^2 – 4x, where k > 0.
3.Find the coordinates of the points of intersection of the line with equation y = kx + 1 and the circle with equation x^2 + y^2 = 9
4. A farmer has a straight, fenced road along the boundary of his property. He wishes to fence an enclosure and has enough materials to erect 500 m of fence. What would be the dimensions to enclose the largest possible rectangular area, assuming that he uses the existing boundary fence as one of the sides?
If you guys could show some working out that would be great :D. Thanks.
Post by: Bad Student on March 20, 2013, 09:11:43 pm
Hey Guys, I've got these three questions that I'd like some help with if anyone's got time.
1.For the quadratic with rule y = 2x^2 + mx+ 4, find the values of m for which there is:
a.one solution
b.two solutions
2.Find the coordinates of the points of intersection of the line with equation y = x + k and the parabola with equation y = x^2 – 4x, where k > 0.
3.Find the coordinates of the points of intersection of the line with equation y = kx + 1 and the circle with equation x^2 + y^2 = 9
If you guys could show some working out that would be great :D. Thanks.
For question 1a, find the discriminant and equate it to 0 and solve it for m.
For question 1b, find the discriminant and solve for m when the discriminant is greater than 0.
To solve question 2, do this
x + k = x^2 – 4x then rearrange it so its all on one side and use the quadratic formula.
Post by: Cammmeron! on March 20, 2013, 09:16:21 pm
Post by: Bad Student on March 20, 2013, 09:23:21 pm
| |
x | | x
| |
---------------------------
500 - 2x
Let the length of each side equal x. The third side will have to equal 500 - x.
Therefore the area of the enclosure will be
After this, you differentiate it, make the derivative equal 0 and solve for x.
Then you sub x back into the area equation to find the maximum area.
Post by: Cammmeron! on March 20, 2013, 09:24:37 pm
Post by: Bad Student on March 20, 2013, 09:25:37 pm
Post by: Daenerys Targaryen on March 20, 2013, 09:26:33 pm
Haha im hilarious.
Post by: Cammmeron! on March 20, 2013, 09:34:03 pm
Post by: Bad Student on March 20, 2013, 09:42:58 pm
Then you expand the left side and move everything onto one side.
Then you use the quadratic formula to solve for x.
And sub the x values back into the first equation to find the y coodinate.
Post by: Bad Student on March 20, 2013, 09:59:05 pm
That is a Terrible Box
Haha im hilarious.
Haha I get it now.
Post by: Conic on March 21, 2013, 04:45:08 pm
------------------------------------
| |
x | | x
| |
---------------------------
500 - 2x
Let the length of each side equal x. The third side will have to equal 500 - x.
Therefore the area of the enclosure will be
After this, you differentiate it, make the derivative equal 0 and solve for x.
Then you sub x back into the area equation to find the maximum area.
At this point of the year you should be using