How do you find the axis of symmetry, vertex and x intercepts for #y=x^2-5x+6#?

1 Answer
Dec 13, 2017

This function is not symmetric to any of the axis.
The vertex is #(5/2,-1/4)#
The x intercepts are #(2,0)# and #(3,0)#

Explanation:

First, to find the axis of symmetry, we need to find whether this function is even or odd.

A function is even if #f(x)=f(-x)#
A function is odd if #-f(x)=f(-x)#

If a function is even, it is symmetric to the y axis. If it is odd, it is symmetric to the origin. Also, it is possible for a function to be neither.

Now, we test the possibilities.
If #f(x)=x^2-5x+6#, then #f(-x)=x^2+5x+6# and #-f(x)=-x^2+5x-6#

We see that none of them are equal to each other. Therefore, this function is not symmetric to any axis.

To find the vertex of a quadratic function, we need to find the #(h,k)#. To do this, remember that #h=(-b)/(2a)# and #k# can be found if you plug #h# into the function. This is a useful trick if the function is in the form of #ax^2+bx+c=y#

Therefore, #h=5/2#

Plugging this into our equation, we get #k=-1/4#

The vertex is #(5/2,-1/4)#

To find the x intercepts, we need to find the zeros of this function.Zeros of the function are the values of #x# that when put into the function, results #y=0#.

Now, we need to solve the equation #x^2-5x+6=0#

We can factor this to get #(x-2)(x-3)=0#

This means that the x intercepts are #(2,0)# and #(3,0)#