Slope of a quadratic function at its roots.

Question

I've realised that every quadratic function with two real roots has slope $-1$ and $1$ at its roots. How do you prove that? Why $1/-1$?

For example: $x^2 - 5x + 6$ has roots = $2$ and $3$. The derivative of the function: $2x - 5$ gives me $1$ and $-1$ when I put $x = 3$ and $2$.

Answer 1

We have that in general

$$f(x)=a(x-s)(x-t) \implies f'(x)=a(x-t)+a(x-s)$$

then

$$f'(t)=a(t-s) \quad f'(s)=a(s-t)$$

therefore your claim holds only for special cases but it is true that in general

$$f'(t)=-f'(s)$$

which depends by the symmetry of the quadratic function.

Answer 2

This is not always true as can be see from the examples in the comments.

Consider the quadratic $ax^2+bx+c$, then it has derivative $2ax+b$. So we have $2ax+b=1$ when $x=\frac{1-b}{2a}$ and $2ax+b=-1$ when $x=\frac{-1-b}{2a}$.

Thats is the derivative of the quadratic gives $1$ and $-1$ when the roots of the quadratic are $\frac{1-b}{2a}$ and $\frac{-1-b}{2a}$ respectively.

In your example we have $x^5-5x+6$, where $b=-5$ and $a=1$. Since $\frac{1-b}{2a}=\frac{6}{2}=3$ and $\frac{-1-b}{2a}=\frac{4}{2}=2$ are roots of the quadratic, then it follows that the slope at these points are $1$ and $-1$.

Answer 3

Geometrically speaking the expression $y=ax^2+bx+c$ is a parabola with a vertical axis.

So it is easy to see that, when the parabola crosses the $x$ axis (zeros are real), then the tangent lines are symmetrical, and the derivatives antisymmetric as per @user 's answer