Question

find the equation of the normal to y=(2x-1)(3x+5)at the point (1,8). give your answer in the form ax+by+c=0, where a,b and c are integers

Answer 1

did you find the derivative yet?
once you do, set it equal to 13 (the gradient or slope)
solve for x (2 values)

then use x in the original to find the y so you get (x,y) for the answer

Answer 2

yh i got 12x+7

Answer 3

now? there are 3 terms with x in
y=2x^3-5x^2+9x-1

term by term what do you get for the derivative?

Answer 4

wait sorry i typed up the wrong question

Answer 5

i've edited it now

Answer 6

yes, I get dy/dx = 12x+7

(two ways to do it: use the product rule, or multiply it out then do term by term)

Then want
equation of the normal to y=(2x-1)(3x+5)at the point (1,8).

normal means perpendicular. The slope of the tangent is dy/dx evaluated at x=1
the slope of the normal is -1/m

first, find dy/dx when x=1

Answer 7

19

Answer 8

-1/19

Answer 9

now what is the slope of the normal line? -1/19

last step
y= mx + b with m= -1/19 x=1 and y=8 and solve for b

Answer 10

152/19?

Answer 11

I get 153/19

Answer 12

oh right yh sorry i forgot to add on 1

Answer 13

8+ 1/19 = (152+1)/19 = 153/19

not quite done, they want
ax+by+c=0, where a,b and c are integers

we have
y= (-1/19)x + 153/19
mult by 19
19y = -x +153
move the stuff from the right to the left side

Answer 14

x+19y-153=0?