Find the equation for the normal line to f at the point (-4, f(-4)).
f(x)=-2x+3x^2

Question

anonymous · Answer

find the derivative.
plug in -4 to get the slope
normal means line is perpendicular so slope is the negative reciprocal
find the equation for the line with that slope through the point given

anonymous · Answer

want to do it?

amistre64 · Answer

y = $-\frac{1}{f'(x)}(x+4)+f(x)$

amistre64 · Answer

i spose f'(-4) and f(-4) are better markers in it :)

anonymous · Answer

guru ! should be $$+f(-4)$$ why am i always cleaning up after...

amistre64 · Answer

i caught it :)

anonymous · Answer

i know i just wanted to crow.  back to the dungeon...

anonymous · Answer

haha! what is the question is telling you to take the derivative? is the line a tangent?

anonymous · Answer

you got this or you need help?

anonymous · Answer

no line is not tangent.  "normal" means perpendicular

anonymous · Answer

lets do it.

anonymous · Answer

$$f(x)=3x^2-2x$$
$$f(-4)=56$$ so the point is (-4,56)

anonymous · Answer

ok, gotcha

anonymous · Answer

$$f'(x)=6x-2$$
$$f'(-4)=-26$$ so slope of tangent line is -36

anonymous · Answer

slope of "normal" line is 
$$\frac{1}{26}$$

anonymous · Answer

you have the slope, you have the point, so find the equation of the line via 
$$y-56=\frac{1}{26}(x+4)$$

anonymous · Answer

or you can write 
$$y-56=\frac{1}{26}x+\frac{2}{3}$$

anonymous · Answer

one sec, Im going through the steps

anonymous · Answer

ok i will wait.  it is now a problem of finding the equation of a line given a slope and a point

amistre64 · Answer

$$y = -\frac{1}{f'(-4)}(x+4)+f(-4)$$

anonymous · Answer

ok, so the derivative is used to calc the slope?

anonymous · Answer

yes.  the derivative itself is not a slope, but it is the formula for one.  so for example you know the derivative of 
$$f(x)=x^2$$ is$$f'(x)=2x$$ this means the slope of the line tangent to 
$$y=x^2$$ at (3,9) is 6 etc

anonymous · Answer

ok, gotcha

anonymous · Answer

but of course in this case finding the slope was one step on the way to finding the slope of the perpendicular line

anonymous · Answer

we got a slope of -26 so the perpendicular line has slope 
$$\frac{1}{26}$$

anonymous · Answer

understand- thanks, I just did another one

anonymous · Answer

good!

anonymous · Answer

where does the (x+4) come from>
In my country we always finished with y=mx+c

anonymous · Answer

thank!

anonymous · Answer

nothing comes up

anonymous · Answer

trying to find the example we did earlier on today where we had x-4/absolute(x-4)
and we had to clac left and and right hand side where x=-4

amistre64 · Answer

there are a lot of different 'forms' that an equation of a line can take; y = mx +c is just one of them

amistre64 · Answer

y = mx -mx0 + y0 is another where your 'c' values = (-mx0+y0)