Tangent question

[rhinwarr]

Hi, could anyone help me with these questions please?

1) Find p such that 2x - y = 1 is a tangent to the parabola y = p(x+0.5)^2 - 1

2) Given f(x) = x^2 - 1, find p such that y = x - 2 is a tangent to y = -f(x-p) - 1

3) Given f(x) = x^2 - 1, find the p values such that y = f(x+2) and y = -f(x-p) touch at a point

Thanks

[Nato]

Hi, could anyone help me with these questions please?

1) Find p such that 2x - y = 1 is a tangent to the parabola y = p(x+0.5)^2 - 1

2) Given f(x) = x^2 - 1, find p such that y = x - 2 is a tangent to y = -f(x-p) - 1

3) Given f(x) = x^2 - 1, find the p values such that y = f(x+2) and y = -f(x-p) touch at a point

Thanks

for question 1 i got p=1
what i did was expand and differentiate y= p(x+0.5)^2-1
>y' = 2px+p
> we know that the line y=2x-1 is a tangent to this curve and the gradient of this line is 2.
>this means at the point y=2x-1 and y = p(x+0.5)^2-1 intersect each other (note: it's only at one points as it is a tangent), the gradient at that specific point on the curve will be the same as the gradient of y=2x-1, namely 2.
>so by differentiating y=p(x+0.5)^2-1 we get 2px+p.
>now we want to make it equal 2 (let the derivative equal to 2 so we can find the specific x points this occurs at)
>solve this 2px+p=2, where x= (2-p)/2p. so we know at the point 2-p/2p on the curve the gradient is equal to zero. now we have to find the y-coordinate. you can sub this into y=2x-1 (as it will be the same for the cure as well) and get (2-2p)/p.
>so the info we have now. at the point (2-p)/(2p)[x-value],(2-2p)/(p)[y-value] is where the tangent occurs. now to find the value of p substitute this into y= p(x+0.5)^2-1 to get the value.
apart from being lazy, you should give a shot at two and three. just from looking at them they carry pretty similar processes.

i hope what i have done here is correct, and if so you understand this. good luck

[SocialRhubarb]

Q1.)




, since the line y=2x-1 must be tangent to a point on the graph, and its gradient is always 2.

Solve for x:

. If we sub these into the original equations:



Solving now for p:



This shouldn't be too hard and you end up with p=1, as above.

[Nato]

Q1.)




, since the line y=2x-1 must be tangent to a point on the graph, and its gradient is always 2.

Solve for x:

. If we sub these into the original equations:



Solving now for p:



This shouldn't be too hard and you end up with p=1, as above.

i should have used latex lol. this is really clear to understand.