How do I convert the equation f(x)=x^2+2/5x−1 to vertex form?

Jul 29, 2015

$\textcolor{red}{f \left(x\right) = {\left(x + \frac{1}{5}\right)}^{2} - \frac{26}{25}}$

Explanation:

The vertex form of a quadratic is given by y = a(x – h)^2 + k, where ($h , k$) is the vertex.

The "$a$" in the vertex form is the same "$a$" as in $y = a {x}^{2} + b x + c$.

$f \left(x\right) = {x}^{2} + \frac{2}{5} x - 1$

We convert to the "vertex form" by completing the square.

Step 1. Move the constant to the other side.

$f \left(x\right) + 1 = {x}^{2} + \frac{2}{5} x$

Step 2. Square the coefficient of $x$ and divide by 4.

${\left(\frac{2}{5}\right)}^{2} / 4 = \frac{\frac{4}{25}}{4} = \frac{1}{25}$

Step 3. Add this value to each side

$f \left(x\right) + 1 + \frac{1}{25} = {x}^{2} + \frac{2}{5} x + \frac{1}{25}$

Step 4. Combine terms.

$f \left(x\right) + \frac{26}{25} = {x}^{2} + \frac{2}{5} x + \frac{1}{25}$

Step 5. Express the right hand side as a square.

$f \left(x\right) + \frac{26}{25} = {\left(x + \frac{1}{5}\right)}^{2}$

Step 5. Isolate $f \left(x\right)$.

$f \left(x\right) = {\left(x + \frac{1}{5}\right)}^{2} - \frac{26}{25}$

The equation is now in vertex form.

y = a(x – h)^2 + k, where ($h , k$) is the vertex.

$h = - \frac{1}{5}$ and $k = - \frac{26}{25}$, so the vertex is at ($- \frac{1}{5} , - \frac{26}{25}$) 