Question

find equations of both lines through the point (2, -3) that are tangent to the parabola y=x^2+x

Answer 1

@Mertsj can you help plz

Answer 2

Take the first derivative of y = x^2+x

Answer 3

okay then?

Answer 4

Replace x with 2 to find the slope of the tangent.

Answer 5

so the slope is 5?

Answer 6

Oh. I don't think that is right because the given point is not on the parabola.

Answer 7

i have to find smaller sloper and larger slope

Answer 8

slope*

Answer 9

I have to think about this.

Answer 10

the answer is -x-1 (smaller slope)
and 11x-25 (larger slope)

Answer 11

but i don't understand how

Answer 12

@Mertsj there?

Answer 13

ok. pick a point on the parabola. If the x value is x, then y=x^2+x so the point(x,x^2+x) is on the parabola. Find the slope from the given point (2,-3) to (x,x^2+x)

Answer 14

can i take a point 2 on the parabola?

Answer 15

That slope is:
\[\frac{x^2+x-3}{x-2}\]

Answer 16

But the slope of the parabola is given by 2x+1 so set the two slopes equal.

Answer 17

\[\frac{x^2+x-3}{x-2}=2x+1\]

Answer 18

solve that equation.

Answer 19

okay

Answer 20

I made a sign error. In the slope, it should be - (-3)

Answer 21

\[\frac{x^2+x+3}{x-2}=2x+1\]

Answer 22

i dont know if im right but i got x=-1 and x=5

Answer 23

\[x^2+x+3=2x^2-3x-2\]
x^2-4x-5=0
(x-5)(x+1)=0
x=5, x=-1

Answer 24

Do you know what to do now?

Answer 25

find y?

Answer 26

yes

Answer 27

okay give me a sec

Answer 28

umm ok so i got (-1, 0) and (5, 30)

Answer 29

Me too.

Answer 30

now how do i find the smaller and larger slope?

Answer 31

The points you just found are the points where the tangents from (2,-3) to the parabola intersect the parabola.
So now you job is to find the equations of the two lines.
One of the lines goes through (5,30) and (2,-3)
The other line goes through (-1,0) and (2,-3)
The easy way to find the slopes of the two lines is to use the slope function which is the first derivative.
y'=m=2x+1
If x = 5, m = 11
If x = -1, m = -1
So now our job is easy.
Find the equation of the line whose slope is 11 and contains the point (2,-3)
Find the equation of the line whose slope is -1 and contains the point (2,-3)

Answer 32

\[y+3=11(x-2)\]
\[y+3=11x-22\]
\[y=11x-25\]
And:
\[y+3=-1(x-2)\]
\[y+3=-x+2\]
\[y=-x-1\]

Answer 33

Thanks a million!!

Answer 34

yw

Answer 35

The graphical result:

Answer 36

Thanks

Answer 37

could you also help me differentiate 5e^x rootx

Answer 38

Chain rule and power rule.

Answer 39

\[5e ^{x}\sqrt{x}\]

Answer 40

\[5e ^{x}((1/2)x ^{-1/2})+ \sqrt{x} (5e ^{x})\]

Answer 41

this doesn't seem right to me

Answer 42

Nevertheless, that is the correct answer.

Answer 43

\[(1/2)x ^{-1/2}+\sqrt{x}\]

Answer 44

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=Integrate%20(5%20E%5Ex)%2F(2%20Sqrt%5Bx%5D)%20%2B%205%20E%5Ex%20Sqrt%5Bx%5D

Answer 45

It is right, however.

Answer 46

yeah i got it now

Answer 47

Differentiate (x+9)/(x^3+x-4)

Answer 48

Please help me out

Answer 49

\[-(2x^3+27x^2+13)/(x^3+x-4)\]

Answer 50

i got that

Answer 51

need to square the denominator

Answer 52

appears correct other than that slip

Answer 53

another question i'm stuck on

Answer 54

For what values of x does the graph of f have a horizontal tangent (use n as integer variable) Enter answer as a comma separated list.

f(x)=x+2sinx

Answer 55

anybody?

Answer 56

Generally, if medals have already been given out on a given problem, I think it likely that some people won't bother looking at it, presumably feeling that it is unlikely they will get one. So unless there is some connection between the problems, in my opinion you are better off posting a fresh question, rather than continuing on. Some people may also feel that if they are going to answer a bunch of questions, they maybe ought to get more than one medal, too.