Help with unit 2 Calculus please!

[sally baker]

and last question.

 
For the curve with equation y=x^2+3 show that y=2ax-a^2+3 is the equation of the tangent at the point (a,a^+3)
hence find the coordinates of the two points on the curve , the tangent of which pass through the point(2,6)

Thanks alot guys

[nbhindi]

8 a) find the equation of the tangent to the curve y=x^3-9x^2+20x-8 at the point(1,4)

a) Let f(x)=y=x^3-9x^2+20x-8 ---->f'(x)=3x^2-18x+20
=>f'(1)=5
.: m=5----> y=5x + c...(1)
Sub Point (1,4) into equ (1)
4 = 5 +c
.: c= -1
.: y=5x -1

[Lasercookie]

8 a) find the equation of the tangent to the curve y=x^3-9x^2+20x-8 at the point(1,4)
b) at what points of the curve is the tangent parallel to the line 4x+y-3=0

I don't mean any offence here, just trying to help you understand this stuff better, but I think you need to look back at the theory. The questions you're asking for help with are kind of starting to be pretty similar. Just doing questions over and over and not learning much isn't going to help - go back to the theory and try to get your head around it. I reckon the Khan Academy videos are top notch for explaining calculus - I strongly suggest you take a look at them.

8a. The question is asking for the equation of the derivative when x=1
happens to equal 4 when x=1 (1,4)
Find dy/dx ()
Sub in x=1

8b. Parallel means the gradients will be equal.
, rearrange to y=mx+c and find the gradient.
Use the derivative of the curve you found in part a and set it equal to the gradient of the line 4x+y-3=0
Solve for x.
Sub this x co-ordinate into the original curve () to find the y co-ordinate.

[nbhindi]

b) at what points of the curve is the tangent parallel to the line 4x+y-3=0

Rearrange equ. ---> y=-4x+3
.: Let f'(x)=-4
-4 = 3x^2-18x+20
0 = 3x^2-18x+24
0 = 3(x-2)(x-4)
.:x=2,4
Sub x=2 into f(x) ----> coordinates= (2,4)
Sub x=4 into f(x) ----> coordinates= (4,-8)

Hope i haven't made a silly error 

[sally baker]

thanks however the answer says  (-4,-56) i think the book is wrong :/

[nbhindi]

For the curve with equation y=x^2+3 show that y=2ax-a^2+3 is the equation of the tangent at the point (a,a^+3)
hence find the coordinates of the two points on the curve , the tangent of which pass through the point(2,6)

Can't be bothered doing it so here are the steps....pm me if u don't understand or need help   
- Differentiate
- Sub x = a into dy/dx---> this will give u gradient of tangent
- Sub coordinates into tangent equ. to solve for "c"

I won't spoil the fun and let u solve part b It's just the same steps as other questions.

[nbhindi]

thanks however the answer says  (-4,-56) i think the book is wrong :/

Yeah i think it's wrong...btw what book r u using...my year 10 methods textbook was riddled with errors
 

[sally baker]

sorry can you guys re explain part be more clearly i dont get it part a The graident is  but for part be is 4 parallel lines have same gradient,.

[paulsterio]

sorry can you guys re explain part be more clearly i dont get it part a The graident is  but for part be is 4 parallel lines have same gradient,.

i think you should have a look at the co-ordinate geometry chapter where they talk about basically parallel lines having the same gradient and perpendicular lines having gradients that multiply to give -1
this does come up again and again even in year 12 methods - basic co-ordinate geometry sort of stuff

and what exactly don't you understand about the question? i can try to help