Find value of a, b, c

[sadness]

Hi! I'm completely new here and I'm hopeless at Calc. Please help me

The curves with equations y = 4/x +2 and f(x) =ax^2 +bx+c have the following properties.
properties:
(i) There is a common point where x = 2;
(ii) There is a common tangent where x = 2;
(iii) Both curves pass through the point (1,6)

Determine the values of a, b and c

Work
i) y = 4/(2) + 2 = 4
f(2) = 4a + 2b +c
ii) y' = -4x^-2
y'(2) = -1
f'(x) = 2ax + b
f''(x)=4a +b
iii) (6) = 4/(1) +2
6= a+b+c

So I'm stuck in terms of what to do in what I solved so far

 

[Deleted member 4993]

Hi! I'm completely new here and I'm hopeless at Calc. Please help me

The curves with equations y = 4/x +2 and f(x) =ax^2 +bx+c have the following properties.
properties:
(i) There is a common point where x = 2;
(ii) There is a common tangent where x = 2;
(iii) Both curves pass through the point (1,6)

Determine the values of a, b and c

Work
i) y = 4/(2) + 2 = 4
f(2) = 4a + 2b +c = 4
ii) y' = -4x^-2
y'(2) = -1
f'(x) = 2ax + b
f''(x)=4a +b
iii) (6) = 4/(1) +2
6= a+b+c

So I'm stuck in terms of what to do in what I solved so far

You have found the slopes of those lines (y' & f') - those must be equal, for common tangent. So,

2a *(2) + b = -1 → 4a + b = -1

Now you have 3 linear-equations and three unknowns.

4a......+ 2b ......+ 1c = 4
4a......+ 1b ......+ 0c = -1
1a .....+ 1b ......+ 1c = 6

Now solve it through your favorite method....

 

[stapel]

First, please let me say "thank you!" for showing your work so nicely!

I'm hopeless at Calc.

Doesn't look that way. You're on the right track!

i) y = 4/(2) + 2 = 4
f(2) = 4a + 2b +c

This gives you a linear equation in variables a, b, and c: 4a + 2b + 1c = 4

ii) y' = -4x^-2
y'(2) = -1
f'(x) = 2ax + b
f''(x)=4a +b

I'm going to guess that "f"(x)" is meant to say "f'(2)". If so, then that line makes sense, and this portion of your work gives you a second linear equation: 4a + 1b + 0c = -1

iii) (6) = 4/(1) +2
6= a+b+c

This is your third linear equation.

You now have three linear equations in three variables, which is a solveable system:

\(\displaystyle \left[\begin{array}{rrrr}4&2&1&4\\4&1&0&-1\\1&1&1&6\end{array}\right]\)

Solve the system for the values of the coefficient variables a, b, and c.

 

[JeffM]

You seem to be OK with mechanics, but do not seem to understand why you are performing those mechanics

Hi! I'm completely new here and I'm hopeless at Calc. Please help me

The curves with equations y = 4/x +2 and f(x) =ax^2 +bx+c have the following properties.
properties:
(i) There is a common point where x = 2;
(ii) There is a common tangent where x = 2;
(iii) Both curves pass through the point (1,6)

Determine the values of a, b and c

Work
i) y = 4/(2) + 2 = 4 OK
f(2) = 4a + 2b +c Good to here but next \(\displaystyle 4a + 2b + c = 4\) because the two functions are equal at x = 2.
ii) y' = -4x^-2 OK
y'(2) = -1 OK
f'(x) = 2ax + b OK but next \(\displaystyle f'(2) = 2a(2) + b = 4a + b = -1\) because the two tangents are equal at x = 2 and so have the same slope.
f''(x)=4a +b Incorrect but irrelevant because the problem says nothing about second derivatives.
iii) (6) = 4/(1) +2
6= a+b+c

So I'm stuck in terms of what to do in what I solved so far

.So you now have three linear equations in three unknowns. What next?

 

[sadness]

You have found the slopes of those lines (y' & f') - those must be equal, for common tangent. So,

2a *(2) + b = -1 → 4a + b = -1

Now you have 3 linear-equations and three unknowns.

4a......+ 2b ......+ 1c = 4
4a......+ 1b ......+ 0c = -1
1a .....+1b ......+ 1c = 6

Now solve it through your favorite method....

Ah....okay thank you !

 

[sadness]

First, please let me say "thank you!" for showing your work so nicely!


Doesn't look that way. You're on the right track!


This gives you a linear equation in variables a, b, and c: 4a + 2b + 1c = 4


I'm going to guess that "f"(x)" is meant to say "f'(2)". If so, then that line makes sense, and this portion of your work gives you a second linear equation: 4a + 1b + 0c = -1


This is your third linear equation.

You now have three linear equations in three variables, which is a solveable system:

\(\displaystyle \left[\begin{array}{rrrr}4&2&1&4\\4&1&0&-1\\1&1&1&6\end{array}\right]\)

Solve the system for the values of the coefficient variables a, b, and c.

Thank you for the explanation > <!

 

[sadness]

You seem to be OK with mechanics, but do not seem to understand why you are performing those mechanics.So you now have three linear equations in three unknowns. What next?

You solve for a, b and c? Thank you for pointing out my key problem about calculus and my errors > <".

 

[JeffM]

You solve for a, b and c? Thank you for pointing out my key problem about calculus and my errors > <".

You do indeed solve for a, b, and c. I did not mean to be sarcastic, and I apologize if my comment came across that way.