How do you write the quadratic in vertex form given #f(x)=5x^2-3x+1#?

1 Answer
Mar 9, 2018

#f(x)=5(x-3/10)^2+11/20#

Explanation:

Vertex form is defined as:

#f(x)=a(x-h)^2+k#

The way to find the vertex is to use #-b/(2a)# (for the #h# value), and then plug that number in for #x# in the original quadratic (for the #k# value).

Now that we know that, we can solve for our #h# value:

#-b/(2a) => -(-3)/(2(5)) => 3/10#

Now that we have our #h# value, we can plug it in for #x# in the quadratic:

#5(3/10)^2-3(3/10)+1#

#=> 5(9/100)-9/10+1 #

#=> 45/100-9/10+1 #

#=> 45/100-90/100+100/100 #

#=> -45/100+100/100 #

#=> 55/100 #

#=> 11/20 #

Now we have our #k# value! We can now switch the quadratic from standard form, to vertex form:

#f(x)=5(x-3/10)^2+11/20#