Find the equation of the tangent to the graph of the function y = 3-x ^ 2 at the point (1; 2)

Let us write the equation of the tangent line in general form:
yk = y0 + y ‘(x0) (x – x0)
By the condition of the problem, x0 = 1, y0 = 2;
Now let’s find the derivative:
y ‘= (3-x ^ 2)’ = -2x;
Consequently:
f ‘(1) = -2 * 1 = -2;
As a result, we have the equation of the tangent to the graph of the function:
yk = y0 + y ‘(x0) (x – x0)
yk = 2 – 2 (x – 1)
or
yk = 4 – 2x;
Answer: y = 4 – 2x.