Quadratics having a common tangent #y = x^2 + ax + b# and #y = cx -x^2# have a common tangent line at the point (1,0), how do you find #a, b# and #c#?

Question

Isabella Ackerman · Accepted Answer

#{a = -3, b = 2, c = 1}#

Calling #f(x) = x^2 + ax + b# and #g(x) = cx -x^2# we have the conditions:

#f(1) = 1 + a + b = 0#
#g(1) = c - 1 = 0#
Now, calling #df(x) = 2x+a# and #dg(x) = c-2x# we have also
#df(1) - dg(1) = 2+a-c+2=0#

Solving for #a,b,c#

#{ ( 1 + a + b = 0), ( c - 1 = 0), (a-c+4=0) :}#

we obtained

#{a = -3, b = 2, c = 1}#

Attached the tangent conics plot.

Axel Alvarez · Answer

undefined To find the values of a, b, and c, we can use the fact that the two quadratic functions have a common tangent line at the point (1,0).

First, let's find the derivatives of the two quadratic functions:
The derivative of y = x^2 + ax + b is dy/dx = 2x + a.
The derivative of y = cx - x^2 is dy/dx = c - 2x.

Since the two functions have a common tangent line, their derivatives must be equal at the point (1,0). Therefore, we can set up the following equation:

2x + a = c - 2x

Now, substitute x = 1 into the equation:

2(1) + a = c - 2(1)
2 + a = c - 2

Since the point (1,0) lies on both tangent lines, we can substitute x = 1 and y = 0 into each quadratic equation:

For y = x^2 + ax + b:
0 = 1^2 + a(1) + b
0 = 1 + a + b

For y = cx - x^2:
0 = c(1) - 1^2
0 = c - 1

Now, we have a system of equations:

2 + a = c - 2
0 = 1 + a + b
0 = c - 1

Solving this system of equations will give us the values of a, b, and c.

HIX Tutor · Answer

undefined Great question! Let’s find the values of $ a $, $ b $, and $ c $ step-by-step.

1. **Understand the Tangent Line**:
   The common tangent line is $ y = x^2 + ax + b $ and $ y = cx - x^2 $. This line touches both curves at the same point.

2. **Find the Point of Tangency**:
   You said the common tangent is at the point $ (1, 0) $. So, let's check if this point lies on both curves.

3. **Substituting Into the First Curve**:
   For the first curve $ y = x^2 + ax + b $:
   $$
   0 = 1^2 + a(1) + b
   $$
   $$
   0 = 1 + a + b \quad \text{(Equation 1)}
   $$

4. **Substituting Into the Second Curve**:
   For the second curve $ y = cx - x^2 $:
   $$
   0 = c(1) - 1^2
   $$
   $$
   0 = c - 1 \quad \text{(Equation 2)}
   $$
   So, $ c = 1 $.

5. **Finding the Derivatives**:
   Now we check the slopes. The slope of the first curve is the derivative:
   $$
   y' = 2x + a
   $$
   At $ x = 1 $:
   $$
   \text{slope}_1 = 2(1) + a = 2 + a
   $$

The slope of the second curve:
   $$
   y' = c - 2x
   $$
   At $ x = 1 $:
   $$
   \text{slope}_2 = 1 - 2(1) = 1 - 2 = -1 \quad \text{(since $ c = 1 $)}
   $$

6. **Setting Slopes Equal**:
   The slopes must be equal because they are the same tangent line:
   $$
   2 + a = -1
   $$
   Solving for $ a $:
   $$
   a = -1 - 2 = -3 \quad \text{(Equation 3)}
   $$

7. **Finding $ b $**:
   Now, substitute $ a = -3 $ back into Equation 1:
   $$
   0 = 1 - 3 + b
   $$
   $$
   0 = -2 + b \quad \Rightarrow \quad b = 2 \quad \text{(Equation 4)}
   $$

8. **Final Values**:
   Now, we have:
   - $ a = -3 $
   - $ b = 2 $
   - $ c = 1 $

**Summary**:
- $ a = -3 $
- $ b = 2 $
- $ c = 1 $

Does that make sense?