As we know that precalculus problems are used to solve many types of problems such as quadratic functions, rational exponents, linear inequalities, lines, inverse functions, algebraic simplification etc. Now let us take an example to understand how to do precalculus problems:
Assume that we are given a quadratic function as f(x) = ax2 + bx + c.
Here a ≠ 0,
This is a second degree polynomial function, so graph of this second order function will be a Parabola. This parabola will open up if a > 0 and parabola will open down if a < 0.
We can calculate x- coordinates of parabola using formula x = -b / 2a. Value of y- coordinate can be calculated by substituting value of x- coordinates in function f (x).
Coordinates we get are in the form of (h, k) then standard equation of parabola can be defined as y – k = a(x - h)2.
x- intercept of parabola is the solution of Quadratic Equation,
ax2 + bx + c = 0.
Calculate the vertices of parabola y = 2x2 - 8x + 7.
Solution: x- intercepts are x = -b / 2a => - (8) / (2 * 2) = 2.
y- intercepts are y = 2(22) – 8 (2) + 7 = -1.