Question

find the parabola with equation y = ax^2 + bx whose tangent line at (1,1) has equation y = 3x-2

Answer 1

first we find b

Answer 2

@lgbasallote I'd written so much but chrome showed me Aww snapp

Answer 3

AWWWW...we need to report this :(

Answer 4

3x - 2 = 2ax + b

Answer 5

at (1,1)

Answer 6

you took the derivative right?

Answer 7

3= 2ax+b
at (1,1)

Answer 8

yeah

Answer 9

okay so we got 3 -2 = 2a + b

Answer 10

this is substitution of (1,1)?

Answer 11

1 = 2a +b okay so we take one of them as eqn

Answer 12

yep

Answer 13

hey...go slow will you -_- lol

Answer 14

so b = 1- 2a

Answer 15

then..

Answer 16

@moha_10 Derivative of the parabola at (1,1 ) is the slope of the parabola and tangent to it at (1, 1)
We are given \(y=ax^2+bx\)
it's slope at (1, 1)=>(m= 2a+b\)
From the tangent's equation, slope is 3 so
\[3=2a+b\]

Answer 17

and finally sub to main function

Answer 18

wait..what happened...

Answer 19

okay may i made a mistake keep going ash