How do you write y=1.4x^2+5.6x+3 in vertex form?

Sep 19, 2016

$y = 1.4 {\left(x + 2\right)}^{2} - 2.6$

Explanation:

$y = 1.4 {x}^{2} + 5.6 x + 3$ is in the form $y = a {x}^{2} + b x + c$

To find the $x$ coordinate of the vertex, use the formula $x = - \frac{b}{2 a}$

$x = - \frac{b}{2 a} = - \frac{5.6}{2 \cdot 1.4} = - 2$

To find the y coordinate of the vertex, plug $x = - 2$ into the equation.

$y = 1.4 {\left(- 2\right)}^{2} + 5.6 \left(- 2\right) + 3 = - 2.6$

The vertex is $\left(- 2 , - 2.6\right)$

Use the formula for the vertex form of a quadratic.
$y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex.
$y = a {\left(x + 2\right)}^{2} - 2.6$

To find the constant $a$, find a convenient point that satisfies the original equation. Typically, the y intercept is found, because the algebra is simple. In other words, find $y$ when $x = 0$.

$y = 1.4 {\left(0\right)}^{2} + 5.6 \left(0\right) + 3 = 3$
Thus, a point that satisfies the equation is $\left(0 , 3\right)$

Use this point to find $a$ by substituting it into the equation for the vertex.
$3 = a {\left(0 + 2\right)}^{2} - 2.6$
$3 = 4 a - 2.6$
$5.6 = 4 a$
$a = 1.4$

Substituting $1.4$ for $a$ into the vertex form equation gives
$y = 1.4 {\left(x + 2\right)}^{2} - 2.6$

There is another process for converting standard form to vertex form called completing the square. If you need to learn this method, please comment, and I will add it to the answer.