How do you write the quadratic function y=-2x^2+6x-3 in vertex form?

Oct 22, 2017

The vertex form of a quadratic of the form $y = a {x}^{2} + b x + c$ is:

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

where $h = - \frac{b}{2 a}$ and $k = a {h}^{2} + b h + c$

Explanation:

Given: $y = - 2 {x}^{2} + 6 x - 3$

Please observe that $a = - 2 , b = 6 \mathmr{and} c = - 3$

Substitute the value of $a = - 2$ into equation :

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

Compute the value of h:

$h = - \frac{6}{2 \left(- 2\right)}$

$h = \frac{3}{2}$

Substitute the value of $h = \frac{3}{2}$ into equation :

$y = - 2 {\left(x - \frac{3}{2}\right)}^{2} + k \text{ }$

Compute the value of k:

$k = - 2 {\left(\frac{3}{2}\right)}^{2} + 6 \left(\frac{3}{2}\right) - 3$

$k = \frac{3}{2}$

Substitute the value of $k = \frac{3}{2}$ into equation :

$y = - 2 {\left(x - \frac{3}{2}\right)}^{2} + \frac{3}{2} \leftarrow$ this is the vertex form

Oct 22, 2017

y = (-2)((x-(3/2)^2) + (3/2)

Explanation:

Vertex form of quadratic equation $a {x}^{2} + b x + c$ is
$y = a {\left(x - {x}_{s}\right)}^{2} + {y}_{s}$

${x}_{s} = - \frac{b}{2 a}$

${y}_{s} = - \left({b}^{2} / 4 a\right) + c$

Given equation is
$y = - 2 {x}^{2} + 6 x - 3$
$a = - 2 , b = 6 , c = - 3$

${x}_{s} = \frac{- 6}{2 \cdot \left(- 2\right)} = \frac{3}{2}$
${y}_{s} = - {b}^{2} / \left(4 a\right) + c = - \frac{- {6}^{2}}{4 \cdot - 2} + \left(- 3\right) = - \left(- \frac{9}{2}\right) - 3 = \frac{3}{2}$

$y = - 2 \cdot {\left(x - \left(\frac{3}{2}\right)\right)}^{2} + \left(\frac{3}{2}\right)$