Only took me 45 minutes to type this up .... not setting any records here :P

Please try using Stapel sources before using my results. I am trying to practice using latex and it took me approx 45 minutes to type this up (yes I am slow) and I do not want to lose my work; therefore, I am going to post it, but you should try to do the work yourself first.

Vertex = ($\displaystyle \frac{3}{4} , \frac{49}{8}$)

x-int = -1 and $\displaystyle \frac{5}{2}$

y-int = 5

domain is (-inf,+inf)

range is (-inf, $\displaystyle \frac{49}{8}$]

See explanation below to determine how I came to these results.

Completing the Square

$\displaystyle

f(x) = -2x^2+3x+5

$

When completing the square the value of "a" should not be any number other than 1; therefore, factor out the -2

$\displaystyle

-2(x^2-\frac{3}{2}x)+5

$

Now divide $\displaystyle -\frac{3}{2}$ by 2 and square the result. The equation should now look like this.

$\displaystyle

-2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16})+5

$

To move the $\displaystyle -\frac {9}{16} $out of the parenthesis you have to multiply it by -2

$\displaystyle

-2(x^2-\frac{3}{2}x+\frac{9}{16})+5+\frac{9}{8}

$

Factor within the parenthesis and clean up outside the parenthesis. The vector format of your function should look like below

$\displaystyle

-2(x-\frac{3}{4})^2+\frac{49}{8}

$

To determine the y-int take the original function $\displaystyle f(x) = -2x^2+3x+5$ substitute 0 for x and you should get the y-int of 5

To get the x-int take the vertex format of the function $\displaystyle -2(x-\frac{3}{4})^2+\frac{49}{8}$ and substitute 0 for y

$\displaystyle -2(x-\frac{3}{4})^2+\frac{49}{8}=0$

$\displaystyle -2(x-\frac{3}{4})^2=-\frac{49}{8}$

Divide both sides by -2

$\displaystyle (x-\frac{3}{4})^2=\frac{49}{16}$

$\displaystyle \sqrt(x-\frac{3}{4})^2=\sqrt\frac{49}{16}$

$\displaystyle x-\frac{3}{4}=\pm\frac{7}{4}$

$\displaystyle x=\frac{3}{4}\pm\frac{7}{4}$

the x-int = -1 , $\displaystyle \frac{5}{2}$