Given f(x)=2x^2+3x-5 for what values of x is f(x) positive

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

Step-by-step explanation:

we have

This is a vertical parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The coordinates of the vertex is the point (h,k)

step 1

Find the vertex of the quadratic function

Factor the leading coefficient 2

Complete the square

Rewrite as perfect squares

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

step 2

Find the x-intercepts (values of x when the value of f(x) is equal to zero)

For f(x)=0

take the square root both sides

therefore

The function f(x) is negative in the interval (-2.5,1)

The function f(x) is positive in the interval (-≠,-2.5) ∪ (1,∞)

see the attached figure to better understand the problem

step-by-step explanation: