# Question 7e641

Apr 12, 2017

$f \left(x\right) = - 2 {\left(x - \frac{5}{2}\right)}^{2} + \frac{15}{2}$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h ,k ) are the coordinates of the vertex and a is a constant.

$\text{to express " f(x)=-2x^2+10x-5" in this form}$

$\text{Use the method of "color(blue)"completing the square}$

We must ensure that the coefficient of the ${x}^{2}$ term is 1

$f \left(x\right) = - 2 \left({x}^{2} - 5 x\right) - 5 \leftarrow \textcolor{red}{\text{coefficient is now 1}}$

add (1/2" coefficient of x-term" )^2 " to " x^2-5x#

$\text{that is } {\left(- \frac{5}{2}\right)}^{2} = \frac{25}{4}$

Since we are adding a number we don't have we must also subtract it.

$\Rightarrow f \left(x\right) = - 2 \left({x}^{2} - 5 x \textcolor{red}{+ \frac{25}{4}} \textcolor{red}{- \frac{25}{4}}\right) - 5$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right)} = - 2 {\left(x - \frac{5}{2}\right)}^{2} + \frac{25}{2} - 5$

$\textcolor{w h i t e}{\Rightarrow f \left(x\right)} = - 2 {\left(x - \frac{5}{2}\right)}^{2} + \frac{15}{2} \leftarrow \textcolor{red}{\text{ in vertex form}}$