A quadratic function is a function whose equation is a second degree polynomial. Thus, if \(f\) is a quadratic function, then \(f ( x ) = a x^2 + b x + c\) for constants \(a, b\), and \(c\), with \(a\ne0\).

Why will the coefficient \(a\) of \(x^2\) never equal zero for quadratic functions?

See SolutionThe graph of a quadratic function is a parabola.

The parabola opens upwards if \(a > 0\) and opens downward if \(a < 0\).

The graph is symmetric about the vertical line \(x = -\dfrac{b}{2a}\).

The vertex of the graph has \(x\) coordinate \(-\dfrac{b}{2a}\).

The zeros of a function are the input values which result in an output value of zero.

What formula did you learn in algebra which allows you to find the zeros of a quadratic function?

See SolutionThere's another way to find the zeros of a quadratic function when the quadratic expression can be factored. For example, let \(f ( x ) = x^2 - x - 6 = ( x - 3 ) ( x + 2 )\).

Then, clearly, the zeros are \(3\) and \(- 2\), since those are the input values which will result in an output value of zero.

Sketch the graph of the function in exercise 2.2.5. Specify the coordinates of the vertex and the \(x\) and \(y\) intercepts.

See SolutionThere is an alternate way to write the equation of a quadratic function using the technique of completing the square which you learned in algebra.

For example, let \(f ( x ) = - x^2 + 4 x + 5\).

First, subtract the \(5\) from both sides, then divide both sides by the coefficient \(-1\) of \(x^2\) to get

\[ \frac{f ( x ) - 5 }{( - 1 ) }= x^2 - 4 x \]Next, add the square of half the coefficient of \(x\) to each side of the equation to get

\[ \frac{f ( x ) - 5}{ ( - 1 )} + 4 = x^2 - 4 x + 4 \]The expression on the right is now a perfect square--the square of \(( x - 2 )\). Following this process will always result in a perfect square on the right side of the equation.

\[ \dfrac{f ( x ) - 5 }{ ( - 1 )} + 4 = ( x - 2 )^2 \]Next, subtract the \(4\) from each side and multiply both sides by \(- 1\) to get

\[ f ( x ) - 5 = - ( x - 2 )^2 + 4\]Finally, add \(5\) to both sides to get

\[ f ( x ) = - ( x - 2 )^2 + 9 \]The advantage of this form of the function is that we can now see that the graph of \(f\) is just the graph of \(y = - x^2\) shifted \(2\) units to the right and \(9\) units upwards. We can also see that the vertex of the graph is at the point \((2,9)\).

The process of completing the square can be used to re-write a quadratic function in the form \(f ( x ) = a ( x - h )^2 + k\) where \(( h,k )\) is the vertex and the graph is just the graph of \(y = a x^2\) shifted \(h\) units horizontally and \(k\) units vertically.

Complete the square on the function and sketch the graph: \(f ( x ) = 2 x^2 - 4 x + 3 \)

See SolutionGiven a quadratic equation \(ax^2+bx+c=0\) the quadratic formula gives the two solutions

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

You may recall that the quantity \(b^2-4ac\) is called the *discriminant*, because we can discover certain facts about the solutions based upon its value.

- If the discriminant is a perfect square, then the quadratic expression can be factored and the solutions are rational.
- If the discriminant is positive, then there are two real and distinct solutions.
- If the discriminant is zero, then there is only one (so-called double) solution.
- If the discriminant is negative, then there are two
*complex*solutions.

Recall that the square root of a negative number is called an *imaginary* number and that we use the symbol \(i\) to denote \(\sqrt{-1}\). Imaginary numbers do not lie on the real number line, since the square of every real number is either positive or zero, but the square of an imaginary number is always negative.

A *complex number* is a number composed of a real part and an imaginary part. For example, \(z=2+3i\) is a complex number composed of the real part \(2\) and the imaginary part \(3i\). The term `complex' is used in the sense of `composed of more than one part' rather than in the sense of `complicated.' We also use the word `complex' in this sense when we speak of an `apartment complex.'

Consider the quadratic equation \(x^2-4x+13=0\).

The quadratic formula gives the solutions \(x=\dfrac{4\pm\sqrt{16-52}}{2}=\dfrac{4\pm6i}{2}=2\pm3i\) which are two complex solutions.