# 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 $f(x)=x^2+4x-3$.

[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?

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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 $y=2x+b$?

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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]