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I'm stuck on a lot of the questions and it's really bugging me because it feels like ive never learnt any of it before!! please help, all will be appreciated

Qu1. A curve has the equation y = √(3x +11)

The point P on the curve has x coordinate 3

(a)Show that the tangent to the curve at P has the equation

3x - 4√5y +31 =0

I've done that bit but i needed the question in context, i cant do this bit:

(b) The normal to the curve at P crosses the y axis at Q, find the y coordinate of Q in the form k√5.

Qu2. The curve with the equation y=x^(5/2)ln(x/4), x>0 crosses the x axis at the point P.

(a) Write down the coordinates of P

The normal to the curve at P crosses the y axis at the point Q

(b) Find the area of the triangle OPQ where O is the origin

The curve has a stationary point at R

(c) Find the x coordinate of R in exact form.

Qu3. f(x) = e ^ (3x+1) - 2

(a) State the range of f

The curve y=f(x) meets the y axis at the point P and the x axis at the point Q

(b) find the exact coordinates of P and Q

(c) show that the tangent to the curve at P has the equation y=3ex + e-2

(d) Find to 3 sig figures the x coordinate of the point where the tnagent to the curve at P meets the tangent to the curve at Q

Qu1. A curve has the equation y = √(3x +11)

The point P on the curve has x coordinate 3

(a)Show that the tangent to the curve at P has the equation

3x - 4√5y +31 =0

I've done that bit but i needed the question in context, i cant do this bit:

(b) The normal to the curve at P crosses the y axis at Q, find the y coordinate of Q in the form k√5.

Qu2. The curve with the equation y=x^(5/2)ln(x/4), x>0 crosses the x axis at the point P.

(a) Write down the coordinates of P

The normal to the curve at P crosses the y axis at the point Q

(b) Find the area of the triangle OPQ where O is the origin

The curve has a stationary point at R

(c) Find the x coordinate of R in exact form.

Qu3. f(x) = e ^ (3x+1) - 2

(a) State the range of f

The curve y=f(x) meets the y axis at the point P and the x axis at the point Q

(b) find the exact coordinates of P and Q

(c) show that the tangent to the curve at P has the equation y=3ex + e-2

(d) Find to 3 sig figures the x coordinate of the point where the tnagent to the curve at P meets the tangent to the curve at Q

Well 1B is easy. Just substitute in x = 0.

∴

3x - 4√5y +31 = 0

- 4√5y +31 = 0

- 4√5y = -31

y = 31/(4√5)

y = (7.75)√5

I think. Not sure about the last bit. Correct me if I'm wrong.

Haven't got time to do the others, sorry.

[EDIT] Ignore everything I just typed up here ^^^^

Didn't notice the question stated the line was the normal to the tangent.

Well rearranging the equation for P into y = mx + c gives:

3x - 4√5y +31 = 0

4√5y = 3x + 31

y = [3/(4√5)]x + 31/(4√5)

The normal to a line has the gradient -1/m

So the line with Q on it is:

y = -1/[3/(4√5)x] + 31/(4√5)

Then substitute in x=0

Anyway I just realised everything I typed, even if it is right, which is unlikely, is a complete mess. So I'm gonna leave this to someone who knows what they're on about.

∴

3x - 4√5y +31 = 0

- 4√5y +31 = 0

- 4√5y = -31

y = 31/(4√5)

y = (7.75)√5

I think. Not sure about the last bit. Correct me if I'm wrong.

Haven't got time to do the others, sorry.

[EDIT] Ignore everything I just typed up here ^^^^

Didn't notice the question stated the line was the normal to the tangent.

Well rearranging the equation for P into y = mx + c gives:

3x - 4√5y +31 = 0

4√5y = 3x + 31

y = [3/(4√5)]x + 31/(4√5)

The normal to a line has the gradient -1/m

So the line with Q on it is:

y = -1/[3/(4√5)x] + 31/(4√5)

Then substitute in x=0

Anyway I just realised everything I typed, even if it is right, which is unlikely, is a complete mess. So I'm gonna leave this to someone who knows what they're on about.

thanks joe_87, now about the others anyone out there??

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