Question

Question about derivative & tangent line.

Will post question and my idea.

Answer 1

So..This is a backwards operation and I'm kinda stuck halfway.

I know:
m = 5

and slope of parabola is:
y' = 2ax + b
m = 2a + b
5 = 2a + b
b = 5 - 2a

where do i plug this b in?

Answer 2

if i plug it in y = ax^2 + bx
i get:
y = ax^2 - 2ax + 5x
y = x(ax - 2a + 5)

...where does that get me.

Answer 3

please tell me m = 2a + b was just a typo...

Answer 4

You have the point (1, 0),W/c is on the parabola
Thus 0 =a+b

Answer 5

well...thats from substituting x = 1

Answer 6

oh.

Answer 7

I got it.

Answer 8

You have 0 = a+b
5 =2a+b
Solve it, then

Answer 9

you just substitute the b to the original equation of y = ax^2 + bx

so...
y = ax^2 - 2ax + 5x
then substitute the point (1,0)
0 = a - 2a + 5
a = 5

then you use a to get the b which is -5.

thanks anyways