At what angle do these curves cut one another? I'm working on an exercise that asks this: At what angle do the curves $$y = 3.5x^2 + 2$$ and $$y = x^2 - 5x + 9.5$$ cut one another? I have set these equations equal to one another to find two values for x. Namely, $x = 1$ and $x = -3$ as intersections. How should I proceed? Most everyone here is always extremely helpful so I can't thank you all enough in advance for any assistance.
 A: The first step is to find the points of intersection, or at least their $x$-coordinates. You have done that.
We deal with the two points of intersection separately. We show how to work with $x=1$.
The derivative of $3.5x^2+2$ is $7x$. So the tangent line at $x=1$ has slope $7$.
Similarly, the tangent line to the other curve at $x=1$ has slope $-3$.
The angle at which the curves meet is the angle between their tangent lines.
Now there are many ways to proceed. We could use the calculator to find separately the angles the two lines make with the positive $x$-axis, and use the results to find the angle between the two lines.
Or else let these angles be $\alpha$ and $\beta$. Use the formula
$$\tan(\alpha-\beta)=\frac{\tan \alpha-\tan\beta}{1+\tan\alpha\tan\beta}.$$
We really want to find the absolute value of $\tan(\alpha-\beta)$. This is readily calculated, since you know that $\tan\alpha=7$ and $\tan\beta=-3$. 
The arithmetic gives $\tan(\alpha-\beta)=\frac{10}{-20}$. Take the absolute value. We get $\frac{1}{2}$. So our angle is the angle between $0$ and $\frac{\pi}{2}$ whose tangent is $\frac{1}{2}$.
If you are taking several variable calculus, or linear algebra, you will know other ways to find the angle between the two lines. 
Repeat with $x=-3$.
A: Think of the curves as being defined by implicit equations $f(x,y)=y - 3.5x^2 - 2=0$ and $g(x,y)=y - x^2 +5x - 9.5=0$. Then the angle between the curves will be the angle between the normal vectors, which are gradients of the two functions $f$ and $g$.
A: Compute the angles between the vectors $$(1, f'(x_i)), (1, g'(x_i))$$ using the formula
$$\cos(\phi) \cdot \Vert u \Vert_2 \Vert v \Vert_2 = \langle u,v \rangle_2$$
Denoting the standard euclidean norm and inner product on $\mathbb{R}^2$.
A: You know the curves cut themselves at $x=1$ and $x=-3$. Let's consider a general case you might find helpful. Consider two functions $f,g$ that intersect at a point $x=\xi$. 
Consider now the tangent line of $f$ at $x=\xi$. What angles does it make with the $x$ axis? It shouldn't be new news that $\tan\theta=f'(\xi)$. Similarily, the tangent line of $g$ at $x=\xi$ makes an angle $\eta$ with $\tan\eta =g'(\xi)$. So at what angle do these lines cross? Would you be convinced it must be $\rho=\theta-\eta$? You can make a drawing, and of course choose $\theta$ to be the largest angle. But we know that $$\tan(\theta-\eta)=\frac{\tan\theta-\tan\eta}{1+\tan\theta\tan\eta}$$
That is, $$\tan(\rho)=\frac{f'(\xi)-g'(\xi)}{1+f'(\xi)g'(\xi)}$$
You should be careful about the angles you're dealing with so as not to get any "off-set" results!
