Answer:
The minimum value is [tex](-\frac{3}{4},-\frac{49}{8})[/tex] or [tex](-0.75,-6.125)[/tex]
Step-by-step explanation:
we have
[tex]f(x)=2x^{2}+3x-5[/tex]
This is the equation a vertical parabola open upward
The vertex represent a minimum
The general equation in vertex form is
[tex]f(x)=a(x-h)^2+k[/tex]
where
(h,k) is the vertex
Convert the given function in vertex form
[tex]f(x)=2x^{2}+3x-5[/tex]
Factor 2
[tex]f(x)=2(x^{2}+\frac{3}{2}x)-5[/tex]
Complete the square
[tex]f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-5-\frac{9}{8}[/tex]
[tex]f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}[/tex]
Rewrite as perfect squares
[tex]f(x)=2(x+\frac{3}{4})^{2}-\frac{49}{8}[/tex]
The vertex is the point [tex](-\frac{3}{4},-\frac{49}{8})[/tex]
