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

Mar 9, 2018

$f \left(x\right) = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{11}{20}$

#### Explanation:

Vertex form is defined as:

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

The way to find the vertex is to use $- \frac{b}{2 a}$ (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:

$- \frac{b}{2 a} \implies - \frac{- 3}{2 \left(5\right)} \implies \frac{3}{10}$

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

$5 {\left(\frac{3}{10}\right)}^{2} - 3 \left(\frac{3}{10}\right) + 1$

$\implies 5 \left(\frac{9}{100}\right) - \frac{9}{10} + 1$

$\implies \frac{45}{100} - \frac{9}{10} + 1$

$\implies \frac{45}{100} - \frac{90}{100} + \frac{100}{100}$

$\implies - \frac{45}{100} + \frac{100}{100}$

$\implies \frac{55}{100}$

$\implies \frac{11}{20}$

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

$f \left(x\right) = 5 {\left(x - \frac{3}{10}\right)}^{2} + \frac{11}{20}$