Can someone correct my questions and help me please.

  • Thread starter 54088
  • Start date
They just want you to summarize the conversation so far. Please do not provide any additional information or answers.In summary, the line y = x + 2 intersects the curve 4y + 3x^2 = 7 in two places. The coordinates of these points of intersection are (-1, 1) and (-1/3, 5/3). The straight line y = f(x) with gradient 3 intersects the curve 4y + 3x^2 = 7 at one point only. To find f(x), we set the equation equal to 0 and solve for x and y, giving us the equation y = 3x + 17/4. To find the coordinates
  • #1
54088
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The line y = x + 2 intersects the curve 4y + 3x^2 = 7 in two places.

a) Find the coordinates of the two points of intersections.

>Sub. y = x + 2 to 4y + 3x^2 = 7
>4(x+2) + 3x^2 = 7
>4x + 8 + 3x^2 -7 = 0
>3x^2 + 4x -1 = 0
>x = -1 and - 1/3
>y = x + 2
>so y = 1 or 5/3

The straight line y = f(x) has gradient 3 and intersects the curve 4y + 3x^2 = 7 at one point only.

b) Find f(x).

>?

c) Hence find the coordinates of the point of intersection between y = f(x) and 4y + 3x^2 =7

>?

:(
 
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  • #2
to question b)
we know that y = f(x) has gradient 3, so y=3x+m, and 4y+3x^2=7
>4(3x+m)+3x^2=7
>3x^2+12x+4m-7=0
at one point only, so the equation above has only one solution, Δ=0
>m=17/4

q c) is easy ^.^
 
  • #3
Thankyou Very Much! ^^"
 
  • #4
54088 said:
The line y = x + 2 intersects the curve 4y + 3x^2 = 7 in two places.

a) Find the coordinates of the two points of intersections.

>Sub. y = x + 2 to 4y + 3x^2 = 7
>4(x+2) + 3x^2 = 7
>4x + 8 + 3x^2 -7 = 0
>3x^2 + 4x -1 = 0
>x = -1 and - 1/3
>y = x + 2
>so y = 1 or 5/3

The straight line y = f(x) has gradient 3 and intersects the curve 4y + 3x^2 = 7 at one point only.

b) Find f(x).

>?

c) Hence find the coordinates of the point of intersection between y = f(x) and 4y + 3x^2 =7

>?

:(

i think that alright you aint to old or to young to learn more
 
  • #5
mensa said:
to question b)
we know that y = f(x) has gradient 3, so y=3x+m, and 4y+3x^2=7
>4(3x+m)+3x^2=7
>3x^2+12x+4m-7=0
at one point only, so the equation above has only one solution, Δ=0
>m=17/4

q c) is easy ^.^

Do NOT do the OP's work for them. They are required to do their own homework here at the PF.
 

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