Question

find the equation of the tangent aat x=3 to the curve with the equation y=2x^2-3x+2.

Answer 1

did you find the derivative of
y= 2x^2-3x +2

dy/dx will be the slope of the curve, after you sub in x=3 into it

also, when x=3, y= 2*3^2- 3*3 +2 = 11
so you will have a point (3,11) and the slope
use
y= mx+b sub in the slope for m, and replace x,y with 3,11 and solve for b

Answer 2

is the derivative 4x-3+0?

Answer 3

That is the correct derivative.

Answer 4

yes
dy/dx = 4x-3 (people don't bother to write +0)

now find m
m= dy/dx with x=3

Answer 5

how do you find m?

Answer 6

@emzy_777 just put the value of x in dy/dx:)

Answer 7

i.e. at x=3 dy/dx=4(3)-3=12-3=8
so y=mx+c
y=8x+c
it must pass through (3,11)
so
11=8(3)+c
c=11-24=-13
so final equation is y=8x-13:)
@emzy_777

Answer 8

thank you @Aperogalics:)

Answer 9

ur welcome:)

Answer 10

@Aperogalics i just checked the answer and the final equation was 9x-7=16...:/

Answer 11

sorry *9x-y=16

Answer 12

ya my fault
at x=3 dy/dx=4(3)-3=12-3=9 not 8:(
so y=9x+c
11=9(3)+c
c=11-27
c=-16
:)

Answer 13

aha its cool:) thank youuu