Question

find the real numbers A and B if the line 8x-y=3 is tangent to the graph of f(X)= Ax^2+Bx+1 at x=1

Answer 1

Well have you found f' yet?

Answer 2

find f' and find the slope of (8x-y=3)
set them equal you have one equation in terms of A and B.
You know the line (8x-y=3) and f(x)=Ax^2+Bx+1 share at least one common point which is the point of tangency (1,f(1))
You can find f(1) by solving 8(1)-y=3 for y.

Answer 3

okay thanks i will try that

Answer 4

@Mathboy23 how are you doing?

Answer 5

\[f'(1)=slope_{of}(8x-y=3) \\ f(1)=A(1)^2+B(1)+1=solve_{for y when x=1}(8x-y=3)\]
You will have a system of equations to solve in the end

Answer 6

I ended up getting B=0 and A=4, do you know if thats right

Answer 7

ok so what did you get the slope of the line 8x-y=3 was 8?

Answer 8

yes i got 8 for the slope

Answer 9

\[f'(1)=8 \\\]
and when you solve 8x-y=3 for y when x=1 you got y=5?
so you have \[f(1)=5\]

Answer 10

yes

Answer 11

so good so far

Answer 12

what system did you end up solving?

Answer 13

Like what the system of equations look like

Answer 14

A+B+1=5 and 2A+B=8 then i multiplied the first equation by 2 and subtracted the second equation

Answer 15

Awesome
your solution looks great

Answer 16

Do you have any questions?

Answer 17

no, thank you so much for all your help :)

Answer 18

np