Question

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

Answer 1

`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`