Problema Solution
in the xy-plane, the graph of the function f(x) = x2 + bx + 4 does not intersect the x-axis. what is the value of b?
Answer provided by our tutors
the graph of the function f(x) = x^2 + bx + 4 does not intersect the x-axis when the quadratic equation
x^2 + bx + 4 = 0 has no real roots
the happens when the discriminant of the equation is < 0
discriminant = b^2 - 4*4*1
b^2 - 4*4*1 < 0
b^2 - 16 < 0
b^2 < 16
|b| < 4^2
|b| < 4
-4 < b < 4
for example for b = 1 we have f(x) = x^2 + x + 4 and we draw the graph we will see that is doesn't intersect the x-axis
click here to see the graph of f(x) = x^2 + x + 4