Need Hints For Two Problems

[scrum]

That is the first one. I figured what i would do is take the derivative of it, and then plug the points in, so i would have 4 equations and then solve the system for each variable. But I'm not getting the right answer for some reason, which i think may be in the algebra but i'm not 100% sure. I'm just wondering if my approach is right, i know i'll be able to get the rest from there.

This is the second problem. I solve part one to y=-2x+b and then I plug in 4 to both equations for x and gey y=-8+b and y=16a. then i realized that this is getting me nowhere so i restarted, i took the derivative on the y=ax^2, and i got y'=2ax. I put the 4 into that and got y=8a. I don't really know where else to go with it, but any hints would be appreciated.

 

[daon]

For the first one: (sorry just read your entire post. it is the correct approach)

If y has a horizontal tangent at (5,10) and (-5,-12) then y' with the x values plugged in will have to be 0.

\(\displaystyle y' = 3ax^2 + 2bx + c\)

Solve the following system. The first two are using the information about the derivative, the second two are using the two points in the original equation.

\(\displaystyle 3a(5)^2 + 2b(5)+c=0\)
\(\displaystyle 3a(-5)^2 + 2b(-5)+c=0\)
\(\displaystyle a(5)^3+b(5)^2+c(5)+d = 10\)
\(\displaystyle a(-5)^3+b(-5)^2+c(-5)+d = -12\)

You have four unknows and four equations.

 

[scrum]

For the first one: (sorry just read your entire post. it is the correct approach)

If y has a horizontal tangent at (5,10) and (-5,-12) then y' with the x values plugged in will have to be 0.

\(\displaystyle y' = 3ax^2 + 2bx + c\)

Solve the following system. The first two are using the information about the derivative, the second two are using the two points in the original equation.

\(\displaystyle 3a(5)^2 + 2b(5)+c=0\)
\(\displaystyle 3a(-5)^2 + 2b(-5)+c=0\)
\(\displaystyle a(5)^3+b(5)^2+c(5)+d = 10\)
\(\displaystyle a(-5)^3+b(-5)^2+c(-5)+d = -12\)

You have four unknows and four equations.

Thanks. The first coordinate is 5,0 not 5,10, but i get what your saying and i got the systems (and solved them by a computer and by hand) but the answer is coming out wrong. I'm gonna work on this some more.

I have grapher for mac so i'm gonna see if that helps me.

any additional help would be greatly appreciated.

 

[scrum]

my system is

0=125a+25b+5c+d
-12=-125a+25b-5c+d
0=75a+10b+c
-12=75a-10b+c

linear sistem solver:
http://wims.unice.fr/wims/wims.cgi
(I can solve systems but this way is quicker and stops me making random algebra errors)

i solve it and get { a = -18/125, b = 3/5, c = 24/5, d = -21 }.

when i graph it it hits (5,0), but misses (-12,5)

 

[daon]

The first two equations are derivatives. In order for the tangent line to have a zero slope the derivative must be equal to zero at the given x value, not equal to the given y value. Your four equations are:

\(\displaystyle y'(5)=0\)
\(\displaystyle y'(-5)=0\)
\(\displaystyle y(5)=0\)
\(\displaystyle y(-5)=-12\)

Or..
\(\displaystyle 75a+10b+c = 0 \\
75a-10b+c = 0 \\
125a+25b+5c+d =0 \\
-125a+25b-5c+d = -12 \\\)