Answer:
[tex]\boxed{ \ 2(x+\dfrac{3}{4})^2-\dfrac{25}{8}\ }[/tex]
Step-by-step explanation:
we should write this function
[tex]2x^2+3x-2[/tex]
this way [tex]a(x-b)^2+c[/tex]
let s check the first terms in[tex]x^2[/tex] and x
[tex]2x^2+3x[/tex]
this is the beginning of
[tex]2(x+\dfrac{3}{4})^2[/tex]
indeed
[tex]2(x+\dfrac{3}{4})^2=2x^2+3x+2(\dfrac{3}{4})^2=2x^2+3x+\dfrac{9}{8}[/tex]
so we can write that
[tex]2x^2+3x=2(x+\dfrac{3}{4})^2-\dfrac{9}{8}[/tex]
then
[tex]2x^2+3x-2=2(x+\dfrac{3}{4})^2-\dfrac{9}{8}-2=2(x+\dfrac{3}{4})^2-\dfrac{9+16}{8}=2(x+\dfrac{3}{4})^2-\dfrac{25}{8}[/tex]
