Question

Determine the equation of the line that is perpendicular to 2y + x = 3 and tangent to the curve f(x)=x^2-3x

Answer 1

The line perpendicular:
1. Rearrange the equation to \(y=mx+c\) and then find the inverse reciprocal of the gradient. Giving you an equation like\[y = \frac{-1x}{m} + c\] where c is arbitrary.

2. Find the tangent to the curve. Simply put them equal to each other.\[x^2-3x = \frac{-1x}{m} + c\]Solve the equation to find c and plug it in. Let me know if that makes sense?

Answer 2

It really doenst make sense to me, i'm supposed to be using derivatives and understandind the process, not using formulas or such.

Answer 3

the way I solved was y'= -1/2 so the slope of the line I want is 2 (2*(-1/2)=-1) and f'(x)=2=2x-3 so x=5/2 f(x)=-5/4 so I get the line I want, that is: y + 5/4=2(x-5/2)

Answer 4

but I dont know if it's right....

Answer 5

It is right.

Answer 6

what program did u use to plot the curves?

Answer 7

This website : http://www.fooplot.com/

Answer 8

I cant solve the next one can u help me?

Answer 9

I can try

Answer 10

Determine the equation of the line that is tangent to both f=-x^2 and g=1/2 + x^2.

Answer 11

Is
\[g=\frac{ 1 }{2 }+x^2\]
or
\[g=\frac{ 1 }{ 2+x^2 }\]?

Answer 12

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

Answer 13

Well, the derivatives are
f''=-2x and
g'=2x

So we need two points a and b for whose, f'(a) = f'(b).
That means
-2a=2b
a=-b

Here is where it gets complicated, because that ecuation has infinetly solutions.
But it can be done by seeing that those graphs are two opposite parabolas and the distance between their vertexes is 1/2, one obvious value for a is -1/2 and, then b = -1/2.

Then just finf f'(-1/2), and write the ecuantoin
And you can verify that the ecuation requested is:
y=x+1/4

Answer 14

* b= 1/2

Answer 15

why obvious values?

Answer 16

Because the two graphs are simetric with the y axis and are also opposite parabolas, that means , that the distance between the tangential point of each one and the center point in between the parabolas ( in this case the point (0,1/4)) is proportional to thhe separation distance.

If the functions were x^2 and -x^2 the common tangent line is simply y=0, because there's no distance between the vertexes.
If you add a value k to anyone of that functions, then the tangent points distance to the center of the two parabolas will increase by an amount proportional to k.

Of course there is a little bit of testing here to find a proportional value.
But is founded on that asumption

http://mathworld.wolfram.com/Parabola.html

I hope i helped.

Answer 17

thx a lot man, it really helped

Answer 18

You're welcome