Question

a bit of help with geometry problem
Find the slope of the tangent line of y = ax^2 + bx + c at (t, at^2 + bt + c).

Answer 1

Since you’re still in geometry I’m going to teach a non-calculus method.

For a parabola, the tangent line only intersects the function once. At that point, the tangent line and the function have the same slope as well as the same y-value.

For a generic line the equation is y = mx + b. Since we want the slope at (t, at^2 + bt + c) let’s use y = mt + b, making our independent variable t instead of x.

the y-values are equal so

mt + b = at^2 + bt + c

Subtracting mt + b from both sides gives us

at^2 + bt - mt + c - b = 0

Which we can rewrite as

at^2 + (b-m)t + (c - b) = 0

This is a quadratic equation which you can solve using the quadratic formula. Once you have the quadratic formula set up, set the discriminant (the square root part) equal to 0. You can do this because our two functions only intersect once and therefore the quadratic equation only has 1 solution, meaning the discriminant is equal to 0.

From there, solve for m