Be careful using the word "bijective." The quadratic function is not bijective. It is bijective on certain subsets of the domain, however.
You can rewrite f(x) as .
I was just taught a formula to solve bijective functions, its shown below:
a[(x +b/2a)^{2 }- (b^{2 }- 4ac)/4a^{2}]
It was used to solve the following problem:
f(x) = 4x^{2 }- 2x + 3
write f(x) in the form f(x) = a(x-b)^{2} + c, (where x > p if f is invertible)
I would like to know what the formula is and how it can prove the function is bijective.
I'm new so forgive me, if i break the rules. Thankyou.
The formula above is just another representation of a quadratic polynomial ax² + bx + c. It is obtained by completing the square. One way to verify it is to derive it using the instructions in the linked page, but it is easier just to simplify the expression to obtain ax² + bx + c.
This expression is used to derive the quadratic formula. It can be used in studying injectivity since the graph of f(x) = ax² + bx + c is a parabola, so f(x) is not injective, but f(x) is injective when restricted to the left or to the right of the parabola's axis of symmetry. The axis of symmetry of the graph of g(x) = p(x - q)² + r goes through x = q, so f(x) = ax² + bx + c is injective on (-∞, -b/(2a)] and [-b/(2a), ∞). The representation g(x) = p(x - q)² + r also shows immediately that the minimum or maximum of g(x) (depending on the sign of p) is r. This can be used in the study of surjectivity.
A small remark: To solve solve a problem or an equation means to find a solution or equation roots, respectively. One cannot solve a function, but one can prove that a function is bijective.