Question

Question is attached. I need help on part C.

Answer 1

why is there no 'x' in the expression you wrote for the equation of the secant line?

Answer 2

It just says to find y

Answer 3

I meant for part a

Answer 4

Oh I see. I forgot to enter it in but yeah x is after 18.

Answer 5

for c find the slope at x=1 and h=.01
use that and the point (1 , 9*1^2 -8*1 +3)
to find the equation of the line (you have the slope of the line and a point on the line...)

Answer 6

wait so i have to put it in the slop intercept form ?

Answer 7

whoops I meant to say "why is there no 'x' in the expression you wrote for the slope of the secant line?"
but you fixed it already anyway...

Answer 8

sure, if you like...

or just y = mx+b
plugging in m, x and y to find b...
either way.

Answer 9

so, I am not sure what to pug in?

Answer 10

to find 'm' plug in x=1 and h=.01 to your expression for the slope.
now you have m. use that and the point x=1 y = 9(1^2) -8(1) +3

Answer 11

in y= mx +b to find 'b'

Answer 12

then you're done... you have m and b in y=mx +b

Answer 13

So its y-4=10.09(x-x1)

Answer 14

y-4=10.09x-10.09

Answer 15

yep

Answer 16

thanks @Algebraic!