Question

The curve y=ax^2 + b passes A (3,39) and B (-2,c). The tangent to the curve at A is parallel to the line y=30x-7. find a,b and c and the tangent to the curve at B

Answer 1

Oh this is fun!

Answer 2

@Lachlan1996 Do you know calculus?

Answer 3

If you take the derivative of ax^2+b, what do you get?

Answer 4

yes, did it months and months ago, been on holidays now

Answer 5

uhhh yes but im not sure if b contains an x so you may not be able to remove b when you derive to get 2ax

Answer 6

I think that you are overthinking it... treat b like a constant.

Answer 7

so... y'=2ax... the derivative is a "slope function", so steal the slope from the equation for the tangent line that they give you.

Answer 8

Yes, a and b are just constants

Answer 9

so 30 = 2a(3)

Answer 10

awesome! solve for a

Answer 11

a = 5

Answer 12

Now use that in the original to solve for b given the point (3,39)

Answer 13

ahh yes so then 39 = a(3)^2 + b

Answer 14

thus 39 = 9a + b

Answer 15

yup!

Answer 16

39 = 9(5) + b
39 = 45 + b
b = -6

Answer 17

yes!

Answer 18

now use everything that you know with that last point and the original equation.

Answer 19

okie dokie, therefore to find c y=5x^2 -6
= 5(-2)^2 -6

Answer 20

therefore c =14

Answer 21

fantastic! It was like a puzzle... how fun!

Answer 22

and gradient of B would be y' = 10x therefore y'= 10(-2) = -20
thus y=mx+c
-2=-20(-2)+c

Answer 23

no wait sorry 14 = -20(-2) + c

Answer 24

therefore c=-26 thus the equation for the tangent at B would be y=-20x-26

Answer 25

Thankyou guys you are awesome!

Answer 26

um... a=5, b=-6, and c=14... done!

Answer 27

oh yeah, tangent to the curve at B... y'=2ax=2(5)(-2)=-20

Answer 28

Thankyou for the help mate :) have a great day

Answer 29

You're welcome!

Answer 30

alrighty cheers for that mate, ill close the question now, and i wish you well