f(x)= 4x^2 + 19x - 5

Factor:

(4x - 1) (x + 5)

etc...

The given expression is a quadratic function in the form f(x) = ax^2 + bx + c, where a = 4, b = 19, and c = -5.

To find the x-intercepts (or zeros) of the function, we need to set f(x) = 0 and solve for x. In other words, we are looking for the values of x for which the function intersects or crosses the x-axis.

So, setting f(x) = 0, we get:

4x^2 + 19x - 5 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values a = 4, b = 19, and c = -5 into the formula, we get:

x = (-19 ± √(19^2 - 4*4*(-5))) / (2*4)

Simplifying further:

x = (-19 ± √(361 + 80)) / 8
x = (-19 ± √441) / 8
x = (-19 ± 21) / 8

This gives us two possible solutions:

x1 = (-19 + 21) / 8 = 2 / 8 = 1/4
x2 = (-19 - 21) / 8 = -40 / 8 = -5

Therefore, the x-intercepts of the given function f(x) = 4x^2 + 19x - 5 are x = 1/4 and x = -5.