Quadratic Functions

A Quadratic function is a polynomial where the highest power is two. The basic form of this function is: F(X) = ax² + bx + c. Where, ax² is the quadratic term, bx is the linear term and c is the independent term or "constant", which does not depend on the variable, x. The letters a and b are called "coefficients", a being the "leading coefficient". The standard form is F(X) = ax² + bx + c. The x-intercepts of the function are:

$x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ OR $x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$

The independent term is also the y-coordinate of the point of intersection with the y-axis (when X=0, F(X)=C).

A quadratic function has a "vertex" or "turning point", which is the point where the function has either a maximum or minimum value. If a is greater than zero, then there will be a minimum and the curve will be concave. If a is less than zero, then there will be a maximum and the curve will be convex. If a = 0, then we have a linear function rather than a quadratic function.The x-coordinate of the vertex is $x=-{\frac {b}{2a}}$ The y-coordinate of the vertex is $F(-{\frac {b}{2a}})$

The general form of a quadratic equation is actually F(X) = ax² + bxy + cy² + dx + ey + f = 0, which can take many shapes including circles, ellipses and parabolas, but in most Western high schools, quadratic equation refers only to those of the form F(X) = ax² + bx + c, which forms a parabola.

Deriving the Quadratic Equation

The solutions to the general-form quadratic function $ax^{2}+bx+c=0$ can be given by a simple equation called the quadratic equation. To solve this equation, recall the completed square form of the quadratic equation derived in the previous section:

$y=a\left(x+{\frac {b}{2a}}\right)^{2}+c-{\frac {b^{2}}{4a}}$

In this case, $y=0$ since we're looking for the root of this function. To solve, first subtract c and divide by a:

$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}$

Take the (plus and minus) square root of both sides to obtain:

$x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}$

Subtracting ${\frac {b}{2a}}$ from both sides:

$x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}$

This is the solution but it's in an inconvenient form. Let's rationalize the denominator of the square root:

${\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}={\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}={\frac {\sqrt {b^{2}-4ac}}{2|a|}}=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}$

Now, adding the fractions, the final version of the quadratic formula is:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

This formula is very useful, and it is suggested that the students memorize it as soon as they can.

Resources

• Quadratic Functions, Book Chapters
• Guided Notes