# Vertex and Equation of Axis of Symmetry of Quadratic Functions

Given a quadratic function $f(x)=ax^2+bx+c$ the vertex and equation of axis of symmetry can be calculated by completing the square of this function. But the derivation will be left to your books and other resources. I will just give the formula how to solve this problem easily.

Definition:

Vertex – the vertex of a quadratic function is the turning point of the graph. This is the point at which the quadratic function changes its direction.

Axis of Symmetry – This is the line that divides the graph of quadratic function into two equal parts.

Formula:

Given a quadratic function $f(x)=ax^2+bx+c$ the vertex $(h,k)$  can be calculated using the following formula.

$h=\displaystyle\frac{-b}{2a}$ ,       $k=\displaystyle\frac{4ac-b^2}{4a}$

The equation of axis of symmetry can be solved using the following equation;

$x=\displaystyle\frac{-b}{2a}$ or in standard form   $2ax+b=0$

Worked Problem 1:

Find the vertex and the axis of symmetry of the function $f(x)=x^2+4x-3$.

Solution:

For vertex $(h,k)$

Given: a=1 , b=4, c=-3

For h:

$h=\displaystyle\frac{-b}{2a}$

$h=\displaystyle\frac{-4}{2(1)}$

$h=-2$

For k:

$k=\displaystyle\frac{4ac-b^2}{4a}$

$k=\displaystyle\frac{4(1)(-3)-(4)^2}{4(1)}$

$k=\displaystyle\frac{4(1)(-3)-(4)^2}{4(1)}$

$k=\displaystyle\frac{-28}{4}$

$k=-7$

For axis of symmetry:

$2ax+b=0$

$2(1)x+(4)=0$

$2x+4=0$ or $x+2=0$

Worked Problem 2:

The vertex of function $f(x)=ax^2+bx+3$ is (1,2). What is/are the value/s of a and b?

Solution:

Using the vertex formula;

$h=\displaystyle\frac{-b}{2a}$

$1=\displaystyle\frac{-b}{2a}$

$b=-2a$ equation 1

$2=\displaystyle\frac{4a(3)-b^2}{4a}$

$8a=12a-b^2$

$b^2=4a$ equation 2.

Using equation 1 and 2:

$b=-2a$    &     $b^2=4a$

Equating a for both equations we have

$\displaystyle\frac{b}{-2}=\displaystyle\frac{b^2}{4}$

$\displaystyle\frac{1}{-2}=\displaystyle\frac{b}{4}$

$b=-2$

Solving for a;

$a=\displaystyle\frac{b}{-2}$

$a=\displaystyle\frac{-2}{-2}$

$a=1$

Worked Problem 3:

For what values of b so that the vertex of the function $f(x)=x^2-3x+2$ is above the line $y=2x+b$?

Solution:

The vertex of $f(x) = x^2 - 3x + 2$ is (3/2, -1/4).
Substitute this to y = 2x + b to get -1/4 = (2)(3/2) + b and b = -13/4
Now the question asked for values of b that will make the vertex above the line, and any value of b less than -13/4 will satisfy, thus the answer is b < -13/4