# Find values of a and b...

• Sep 24th 2007, 08:37 PM
ebonyscythe
Find values of a and b...
I'm really having trouble with these two problems in my homework... I'm just not sure where to start and would appreciate a push in the right direction.

In my class, we've just gotten through the quotient rule for derivatives.

1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2

This is what I did, but my answer did not check out when I graphed it and such:

a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?
So...
a + b = 1 or b = 1-a or a = 1-b

y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
2a(1) + b = 3
So...
2a + b = 3

2a + (1-a) = 3
a = 4

2(1-b) + b = 3
2-b = 3
b = -1

But when I graphed both, the equation of the tangent line didn't lie on (1,1) of the graph y = 4x^2 - x. Where did I go wrong?

2. Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y= -x^2 +6x - 5.

And I don't know where to start for this one...
• Sep 24th 2007, 09:02 PM
Jhevon
Quote:

Originally Posted by ebonyscythe
I'm really having trouble with these two problems in my homework... I'm just not sure where to start and would appreciate a push in the right direction.

In my class, we've just gotten through the quotient rule for derivatives.

1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2

This is what I did, but my answer did not check out when I graphed it and such:

a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?
So...
a + b = 1 or b = 1-a or a = 1-b

y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
2a(1) + b = 3
So...
2a + b = 3

2a + (1-a) = 3
a = 4

your mistake was in this last line. you should subtract 1 from both sides, not add it. a = 2
• Sep 24th 2007, 09:16 PM
ebonyscythe
Thanks, I knew I did something ridiculous... can't believe I didn't catch that one while typing.
• Sep 24th 2007, 09:49 PM
ticbol
1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2

a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?

The Lefthand side is okay, but the RHS is...where did you get that.

The RHS should be 1 only.
Yours came out as 1 also, but that is by chance only.

y = ax^2 +bx
At (1,1),
1 = a(1^2) +b(1)
1 = a +b ----------------------(1)

y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
2a(1) + b = 3
So...
2a + b = 3 -------------------(2)
Since, from (1), b = 1-a, then, in (2),
2a +(1-a) = 3
2a -a = 3 -1
And so, b = 1 -2 = -1 ----------------answer.

Check,
y = ax^2 +bx
y = 2x^2 +(-1)x
y = 2x^2 -x
Is (1,1) a point of the curve?
1 =? 2(1^2) -1
1 =? 1
Yes, so, OK.

---------------------------------------------------------

2. Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y= -x^2 +6x - 5.

Say there is only one common line that is tangent to both curves

On the y = x^2 parabola,
m1 = y' = 2x

On the y = -x^2 +6x -5 parabola,
m2 = y' = -2x +6

It is the same tangent line, so its slope is the same everywhere along the line.
So, m1 = m2
2x = -2x +6
4x = 6
x = 6/4 = 3/2
And so,
m1 = 2x = 2(3/2) = 3
m2 = -2x +6 = -2(3/2) +6 = 3 also.

So the equation of the tangent line is y = 3x +b -----(i)

(i) intersects the 1st parabola at their point of tangency, so, their y's there are the same. Hence,
3x +b = x^2
x^2 -3x -b = 0 -------------------(ii)

Likewise, on the 2nd parabola,
3x +b = -x^2 +6x -5
x^2 -6x +3x +b +5 = 0
x^2 -3x +5 +b = 0 ---------------(iii)

(iii) minus (ii),
5 +b +b = 0
2b = -5
b = -5/2 --------------***

Therefore, the equation of the tangent line is
y = 3x +b -----(i)