Question

Find the equation of the tangent lines to the parabola y=x^2+x that pass through the point (2, -3) . Sketch the curve and the tangents.

Answer 1

tangent line to f(x)=x^2+x at x=a is given by
\[y-f(a)=f'(a)(x-a)\]

Answer 2

So you will need to find f'(x) first

Answer 3

can you do that?

Answer 4

is it y'=2x+1??

Answer 5

yeah
now what is f(a) and f'(a)

Answer 6

\[f(x)=x^2+x \\ \text{ so } \\ f(a)=? \\ \text{ and } f'(x)=2x+1 \\ \text{ so } f'(a)=?\]

Answer 7

f(a)= a^2+a
f'(a)=2a+1

Answer 8

great so we have this so far:
\[y-f(a)=f'(a)(x-a) \\ y-(a^2+a)=(2a+1)(x-a)\]

Answer 9

now we know a point on this line

Answer 10

it was given to us

Answer 11

we wanted the point (2,-3) on this line
so we can find a such that will happen by replacing x with 3 and y with -3 and solving for a

Answer 12

\[y-f(a)=f'(a)(x-a) \\ y-(a^2+a)=(2a+1)(x-a) \\ -3-(a^2+a)=(2a+1)(3-a)\]
see if you solve this for a
it should just be a quadratic

Answer 13

i have a question, should it be x=2, not x=3

Answer 14

yes I'm dyslexic today

Answer 15

actually was one of them negative

Answer 16

x=2 and y=-3

Answer 17

it will be -(a-5)(a-1)
then a=1,5

Answer 18

hmm...

Answer 19

I got similar values
one is just a bit different in sign then one you got

Answer 20

i mean -(a-5)(a+1)
a=5, -1

Answer 21

oops

Answer 22

great work

Answer 23

that is exactly what i got

Answer 24

now we have to a few more things

Answer 25

it asked for equation of tangent line for which it was suppose to go through (2,-3)

Answer 26

\[y-f(a)=f'(a)(x-a) \\ y-(a^2+a)=(2a+1)(x-a) \\ \text{ replace } a \text{ with } 5 \text{ for one possible equation } \\ \text{ then replace } a \text{ with -1 } \text{ for another possible equation }\]

Answer 27

Are you cool with that?

Answer 28

for example I will give you equation:
\[a=5 \\ y-(5^2+5)=(2 \cdot 5+1)(x-5) \\ y-(25+5)=(10+1)(x-5) \\ y-30=11(x-5)\]
this equation for the tangent line at x=5 is in point-slope form

Answer 29

still working on it

Answer 30

now take the same equation I just used and replace a's with -1

Answer 31

I don't know what the required form is
like slope-intercept
or point slope form

Answer 32

i got y=x+1

Answer 33

\[a=-1 \\ y-(a^2+a)=(2a+1)(x-a) \\ y-([-1]^2+[-1])=(2 \cdot [-1]+1)(x-[-1]) \\ y-(1-1)=(-2+1)(x+1) \\ y-0=(-1)(x+1) \\ y-0=-1(x+1) \\ y=-(x+1)\]
check it one more time
just and case I made a mistake

Answer 34

it looks right

Answer 35

your equation or mine

Answer 36

yours, i forgot a negative in mine

Answer 37

ok
just checking to see if you agree with my steps
and also I'm not wearing my glasses lol

Answer 38

do you have any questions on this problem?

Answer 39

you have two answers

Answer 40

how do you show it as "11x-y-25=0" and "x+y+1=0" (this is the answer from my textbook)?

Answer 41

nevermind, i got it. @freckles
Thank you for helping me! :D

Answer 42

alrighty :)

Answer 43

your mention didn't show through for some reason