stationary points (1 Viewer)

[Arithela]

1. The curve y = ax^2 - 6x + b has a stationary point at (3,-2) on it.

Determine the values of a, b.

2. There is a turning point at (1,13) on the curve y = ax^2 + bx + 12. Find a and b.

3. If y = ax^2 + bx + c touches the line y = x at the origin and has a max. turning point where x = 1, find the values of a, b and c.

 

[loser101]

easy.

given y = ax^2-6x-b

differentiate => y'=2ax-6

stationary point we put y'=0 => i.e 2ax-6=0 => solve for x => x=3/a

but we are given we have a stationary point at (3,2), this tells us that y'=0 for x=3 and y=2

Therefore x=3/a=3 => a=1

substitue a=1, x=3, y=-2 into your original equation y=ax^2 - 6x +b

solve for b => b = -2 -(1)(3)^2+6(3) = 7


i.e a =1, b =7

its much shorter than what i have written out...

this problem is testing you know:

Stationary point dy/dx = 0 a for stationary point
How to solve by substition linear equations

 

[independantz]

1)f(x)=ax^2-6x+b
f'(x)= 2ax-6, since there is a stationary point at (3.-2)
f'(3)=6a-6=0 therefore, a=1
Since the y-value at 3 is -2
f(3)=9-18+b=-2
Therefore, by rearranging b=7
i.e. f(x)=x^2-6x+7

2)Is basically the same thing.

3)f(x)=ax^2+ bx + c
f'(x)=2ax+b at x=1 f'(x)=0
i.e.: f'(x)=2a+b=0-(1)
Since the line touches y=x at the origin. then, (o,o)should satisfy the equation
therefore,0= c, by rearranging we get c=0.

umm sorry i don't have enough time at the momnet, i'll finish the rest when ig et back, but you should be able to work it from there.

 

[loser101]

Q2)


given y = ax^2+bx+12 ------------------ (1)

y' = 2ax+b

given y' is stationary pt at (1,13) i.e y'=0 at x=1 and y=13

solve y'=0=2a(1)+b

b = -2a --------------(2)

substitute b=-2a back into the (1)

y=ax^2-2ax+12

we have pt (1,13)

i.e 13=a(1)^2-2a(1)+12

mental gymnastics, a=-1

from eqn (2) b=-2a

b=2


i.e a=-1, b=2

 

[loser101]

Q3)
y = ax^2 + bx + c

points we know on the quadratic eqn (0,0) since it lies on the line y=x at the origin (0,0)

subst (0,0) => c=0

y'=2ax+b

y'=0 at x=1

b=-2a

^^^ this part not necessary

also a<0 since the Quad ftn is a max (i.e concave down)

by inspection and knowledge of quadtracitc ftns

we know the roots are at x=0 and x=2 and the symetry is at x=1 and its concave down

y= -(x)(x-2)

i.e y=-x^2+2x+0

i.e a=-1, b=2, c=0

Testing Knowledge of

what the a, b and c does to the porabola y = ax^2 + bx + c

also what y= (x-x1)(x-x2) means

 

[flaming.robo]

1)f(x)=ax^2-6x+b
f'(x)= 2ax-6, since there is a stationary point at (3.-2)
f'(3)=6a-6=0 therefore, a=1
Since the y-value at 3 is -2
f(3)=9-18+b=-2
Therefore, by rearranging b=7
i.e. f(x)=x^2-6x+7

2)Is basically the same thing.