Question

Find the max/min value of f(x) = 2x2 + 3x - 5

Answer 1

Answer:

The minimum value is $(-\frac{3}{4},-\frac{49}{8})$ or  $(-0.75,-6.125)$

Step-by-step explanation:

we have

$f(x)=2x^{2}+3x-5$

This is the equation a vertical parabola open upward

The vertex represent a minimum

The general equation in vertex form is

$f(x)=a(x-h)^2+k$

where

(h,k) is the vertex

Convert the given function in vertex form

$f(x)=2x^{2}+3x-5$

Factor 2

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

Complete the square

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

$f(x)=2(x^{2}+\frac{3}{2}x+\frac{9}{16})-\frac{49}{8}$

Rewrite as perfect squares

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

The vertex is the point $(-\frac{3}{4},-\frac{49}{8})$