Why does the solution for the derivative of the quadratic function on x value, equals the first coefficient on the linear function?

Question

I have the following function $$() = 2² − 3 − 2$$ and I have to find the angle made by the tangent to this function at $$ =(1, −3)$$ with the −axis.

I know the linear function $$() = + n$$ and $$ =(1, −3)$$ so $$(1) = + = −3$$ and then to find m there is the derivative of the quadratic $$′() = 4 − 3$$ $$′(1) = 4 − 3 = 1 ⟹ m$$

I assumed that because the derivative and the linear function belong to the same quadratic, they will have the same properties, and that's why we use the derivative.

But why m equals 1? Is there some flaw in my reasoning?

Answer 1

The slope of the line $L(a)$ that is tangent to the graph of $f$ at $(a, f(a))$ is $f'(a).$ The slope of the line whose equation is $y=mx+n$ is $m,$ so if this line is $L(a)$ then $m=f'(a).$ In particular if $a=1$ and $f(a)=-3$ and $(a,f(a))=(1,-3)=A$ then $m=f'(a)=f'(1)=1.$ Your reasoning is correct.