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

answer: no, but if it does it's a obvious pattern.

step-by-step explanation:

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!