Let $\varepsilon_1,\varepsilon_2$ be the equations of the tangents to the curve $c: y = 1+x^2$, on the points $x_1 = 2$ and $x_2 = -2$, respectively. Prove that the equations for the two tangents are $\varepsilon_1: y = 4x -3$ and $\varepsilon_2: y = -4x-3\;$.
Unfortunately, I have no hunch or idea on how to begin on this. We've got the two points but how would I go about proving that the given tangent equations are the correct ones?
Source: a sheet of the class's 2012 final exams, I'm slowly trying to solve the exercises in it or find the general layout. I would like to ask for any help or methodologies that will make it easier for me to solve.
The derivative of a function can be interpreted geometrically as the slope of the tangent to the function.
The derivative of $1+x^2$ is $2x$, so substituting the given values for $x_1$ and $x_2$ we obtain that the slope of the tangent line in those points is $4$ and $-4$.
Now you should recall from analytic geometry that the formula for a line passing through the point $(x_0,y_0)$ with slope $m$ is $y-y_0=m(x-x_0)$
For example the line passing through $(2,5)$ (the point with x-coordinate equal to $2$ in your curve) and having slope $m=4$ is given by $y-5=4(x-2)$ that is $y=4x-3$
A similar calculation will show that also the other tangent is correct