sketch the curves y=x^2+1 and y= -x^2 on the same axes. Find the equations of the lines that are tangent to both curves simulaneously

To solve the problem, let's go through the following steps:

### 1. **Sketch the Curves \(y = x^2 + 1\) and \(y = -x^2\):**

- **Curve 1: \( y = x^2 + 1 \)**
  - This is a parabola that opens upwards with its vertex at \((0, 1)\).

- **Curve 2: \( y = -x^2 \)**
  - This is a parabola that opens downwards with its vertex at \((0, 0)\).

The curves intersect the y-axis at \( y = 1 \) for the first curve and \( y = 0 \) for the second curve. Let's proceed to plot these curves.

### 2. **Find the Equations of the Tangent Lines:**

We need to find the equations of the lines that are tangent to both curves simultaneously.

Let the equation of the tangent line be:
\[
y = mx + c
\]

For it to be tangent to \(y = x^2 + 1\), the line should touch the parabola at exactly one point.

Substitute \(y = mx + c\) into \(y = x^2 + 1\):
\[
mx + c = x^2 + 1
\]
Rearrange this into a quadratic equation:
\[
x^2 - mx + (1 - c) = 0
\]
For the line to be tangent, the discriminant of this quadratic equation must be zero:
\[
\text{Discriminant} = (-m)^2 - 4(1)(1 - c) = 0
\]
\[
m^2 - 4 + 4c = 0 \quad \Rightarrow \quad c = \frac{m^2 - 4}{4}
\]

Now, for the line to be tangent to \(y = -x^2\):
\[
mx + c = -x^2
\]
Rearrange this into a quadratic equation:
\[
x^2 + mx + c = 0
\]
Again, for the line to be tangent, the discriminant must be zero:
\[
\text{Discriminant} = m^2 - 4c = 0 \quad \Rightarrow \quad c = \frac{m^2}{4}
\]

### 3. **Equating the Tangent Conditions:**
We need to find values of \(m\) and \(c\) such that the tangents are the same for both curves:
\[
\frac{m^2 - 4}{4} = \frac{m^2}{4}
\]
Simplify to find \(m\):
\[
m^2 - 4 = m^2
\]
This implies that \(4 = 0\), which cannot happen.

Thus, there is no common \(m\) that can satisfy the tangency conditions simultaneously, indicating that there is no single line that can be tangent to both curves simultaneously.

### Conclusion
There are no lines that can be tangent to both \( y = x^2 + 1 \) and \( y = -x^2 \) simultaneously.

Would you like any further details or have any questions?

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**Related Questions:**

1. What is the significance of the discriminant in determining the tangency of a line to a curve?
2. Can you find the points of intersection of the curves \(y = x^2 + 1\) and \(y = -x^2\)?
3. How do the vertices of the parabolas affect the possibility of simultaneous tangency?
4. What are the general forms of the tangent lines for \(y = x^2 + 1\) and \(y = -x^2\)?
5. What would happen if the curves were shifted vertically or horizontally?
6. How does the concavity of the curves affect the tangency conditions?
7. How would you find the tangent lines to both curves if they were translated?
8. Can you find the area enclosed between the curves \(y = x^2 + 1\) and \(y = -x^2\)?

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**Tip:** Always check the discriminant condition for quadratic equations when finding tangent lines, as it determines whether the line touches the curve at one point or intersects it at two points.

Learn how to find tangent lines to the curves y = x^2 + 1 and y = -x^2 simultaneously with a detailed step-by-step solution. Explore the quadratic equations and the discriminant's role in determining tangent conditions. Suitable for advanced high school students and calculus enthusiasts.

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