Question

Find the equation for the line tangent to the graph of the given function at the indicated point.

f(x)=x-x^2 at (-4,-20)

Answer 1

f'(x) = 1- 2x
-4 = 1- 2(-20) + b
-5 = -40 + b
b = 35

equation: y = 36 -2x might need to double check with someone else

Answer 2

thanks for replying! the answer key says its y=9x+16. not sure how to get to that answer

Answer 3

sorry i messed up unfortunately i forgotten my calc, all i know is that derviative will give you an equation tagent to the function but im not exactly sure how to do it for a specific point sorry

Answer 4

no worries but thanks for trying

Answer 5

ok...do you understand what the 1st derivative gives you..?

Answer 6

1-2x?

Answer 7

the 1st derivative gives the general equation of the slope of the tangent.

f'(x) = 1 - 2x

you need to find the slope at the point where x = -4

so evaluate

f'(-4) = 2 - 2(-4)

Answer 8

=10

Answer 9

ok... so the slope of the tangent is m = 10

so you have a point (-4, -20) and a slope m = 10

can you find the equation of the line..?

Answer 10

but you need to check you solution for the value of the slope...

f'(-4) = 1 - 2(-4)

Answer 11

yay i got it! so the slope is 9

y-(-20)=9(x-(-4)

y+20=9x+36
-20 -20
y=9x+16

thanks!!!!! :)

Answer 12

that correct... just remember the derivative is the equation of the slope of the tangent... substitute the x value to get the slope at a specific point...

glad it helped.