# Thread: Find values of a and b...

1. ## 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...

2. 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

3. Thanks, I knew I did something ridiculous... can't believe I didn't catch that one while typing.

4. 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)