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.vertex eqn symm


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.


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.

[toggle title=”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 [/toggle]


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?

[toggle title=”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 [/toggle]


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?

[toggle title=”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[/toggle]

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