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

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.

[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]

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

Latest posts by Dan (see all)

You may also like...

3 Responses

  1. pron best says:

    cjajFH This is really fascinating, You are a very professional blogger. I ave joined your rss feed and sit up for searching for more of your great post. Also, I have shared your site in my social networks!

  2. suba me says:

    tZNKe1 Marvelous, what a weblog it is! This web site provides helpful information to us, keep it up.

  3. 767050 494514I like this web site really significantly, Its a actually good situation to read and get information . 740657

Leave a Reply

Your email address will not be published.