 The solution of quadratic equations is important in many fields, such as engineering, architecture and astronomy. Studies include axis of symmetrymaximum and minimum values roots of quadratic equations and formula, etc.

Using the knowledge from Introduction to Functions, the theory could be expanded further in this topic.

Graph of a Quadratic Function

A quadratic function's graph is a parabola where it is either concave up or concave down. Remember, when the coefficient in front of the ${x}^{2}$ term is positive, the curve would concave up. On the other hand, when the coefficient in front of the ${x}^{2}$ term is negative, the curve would concave down. Therefore there is also an axis of symmetry which divides the parabola in the middle. For example, using the intercept method we could sketch the parabola $y={x}^{2}-4x$ on the number plane. $For\quad the\quad y-intercept,\quad x=0\\ i.e.\quad y={ 0 }^{ 2 }-4(0)\\ \qquad y=0\\ \\ For\quad the\quad x-intercept,\quad y=0\\ i.e.\quad { x }^{ 2 }-4x=0\\ \qquad x(x-4)=0\\ \qquad x=0\qquad or\qquad x-4=0\\ \qquad \qquad \qquad \qquad \quad x=4$ Knowing that the x-intercept lies on the points x=0 and x=4, the middle of the curve would naturally be in the middle.

This means that the axis of symmetry is the equation x=2.

What is the minimum value of the parabola/quadratic function $y={x}^{2}-4x$?

Since we already know that the curve changes direction at x=2, the minimum value should be on that particular point. From the graph, at x=2 the y-value is y=-4. We could substitute the value x=2 to the function to find the value of y. $y={ x }^{ 2 }-4x\\ y={ 2 }^{ 2 }-4(2)\\ y=4-8\\ y=-4$

Otherwise, the axis of symmetry of a quadratic function $y=a{x}^{2}+bx+c$ has the equation: $x=-\cfrac{b}{2a}$ The maximum or minimum value is $f(-\cfrac{b}{2a})$.

Example:

Find the equation of the axis of symmetry and the minimum value of the quadratic function $y={x}^{2}-5x+1$.

The equation of the axis of symmetry is given by: $x=-\cfrac { b }{ 2a } \\ \\ x=-\cfrac { -5 }{ 2(1) } \\ \\ x=\cfrac { 5 }{ 2 } \\ \\ \therefore \quad Equation\quad is\quad x=2\frac { 1 }{ 2 } \\ \\ Minimum\quad value:\\ \\ y={ \left( \cfrac { 5 }{ 2 } \right) }^{ 2 }-5{ \left( \cfrac { 5 }{ 2 } \right) }+1\\ \\ \quad =\cfrac { 25 }{ 4 } -\cfrac { 25 }{ 2 } +1\\ \\ \quad =-5\frac { 1 }{ 4 } \\ \\ \therefore \quad Minimum\quad value\quad is\quad -5\frac { 1 }{ 4 } \\ \\ \qquad \qquad \qquad \qquad \\$

Think: Could you tell without sketching the function $y={x}^{2}-x+5$ if $y={x}^{2}-x+5>0$ for all x? How could you do this?

How could you confirm that $-{x}^{2}+2x-7<0$ for all x without skteching the graph of $f(x)=-{x}^{2}+2x-7$?