How do I convert the equation $f (x) = x^{2} + \frac{2}{5} x - 1$ $f(x)=x^2+2/5x−1$ to vertex form?

Question

3408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-29T18:57:19

$f (x) = {(x + \frac{1}{5})}^{2} - \frac{26}{25}$ $color(red)( f(x) = (x+1/5)^2-26/25)$

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 $y = ax^2 + bx + c$ .

Your equation is

$f(x) = x^2+2/5x-1$

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

Step 1. Move the constant to the other side.

$f(x)+1 = x^2+2/5x$

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

$(2/5)^2/4 = (4/25)/4 = 1/25$

Step 3. Add this value to each side

$f(x)+1+1/25 = x^2+2/5x+1/25$

Step 4. Combine terms.

$f(x)+26/25 = x^2+2/5x+1/25$

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

$f(x)+26/25 = (x+1/5)^2$

Step 5. Isolate $f(x)$ .

$f(x) = (x+1/5)^2-26/25$

The equation is now in vertex form.

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

$h = -1/5$ and $k = -26/25$ , so the vertex is at ( $-1/5,-26/25$ )

Graph

How do I convert the equation f(x)=x^2+2/5x−1f(x)=x2+25x−1f(x)=x^2+2/5x−1 to vertex form?