Question

how do find the equation of the tangent and the equation of the secant of y=x^2 at x=-2 ?

Answer 1

for the secant line you would need two points
for the line tangent at \((-2,4)\) you need a slope, which is given by the derivative evaluated at \(x=-2\)

Answer 2

ive gotten that the slope is -4

Answer 3

the derivative of \(f(x)=x^2\) is \(f'(x)=2x\) and \(f'(-2)=-4\) meaning the slope of your line is \(-4\)

Answer 4

ok good
last step is to use the point slope formula since you have the slope is \(-4\) and the point is \((-2,4)\)

Answer 5

this is for the tangent?

Answer 6

yes

Answer 7

for a "secant" line you are going to need two points on the curve, not one

Answer 8

ok. i got y = -4x - 4.
now for the secant, how do i get two points?

Answer 9

the question is not well posed
you cannot be asked to find the equation of "the" secant line at only one point. you need to be told two points.
is that exactly how the question reads?

Answer 10

hmm. it just says:
"at the indicated point find
a) the slope of the curve
b) the equation of the tangent, and
c) an equation of the normal
wait... so then im guessing "secant" and "normal" arent the same thing... haha.

Answer 11

ooooooooooooooh no

Answer 12

you want the "normal line" that means the line perpendicular to the curve at that point

Answer 13

soooo reciprocal of -4 and find b

Answer 14

but you have done the main part of the work already
slope of tangent line is \(-4\) so slope of perpendicular line is \(\frac{1}{4}\)

Answer 15

careful, not the reciprocal, the negative reciprocal
point is of course the same so it is just another application of the point slope formula

Answer 16

ahhh yes thats what i meant. i got y=(1/4)x + (9/2)
thanks! (:

Answer 17

yw

Answer 18

oh and yes, that is right