Question

For how many integers L do the graphs of y=x^2-3x-4 and y=x^2+Lx intersect at (at least) one non-negative y value?

a) 0
b) 1
c) 2
d) Infinite

Answer 1

There are infinite integers L that allow the graphs y=x^2-3x-4 and y=x^2+Lx to intersect with non-negative y values since, for x>=4, L can be any integer without making y negative.

Step-by-step explanation:

The question asks for the number of integers L that allow the graphs of y=x^2-3x-4 and y=x^2+Lx to intersect where y is non-negative. To find intersections, we set the equations equal to each other:

x^2 - 3x - 4 = x^2 + Lx
-3x - 4 = Lx
L = -3 - (4/x)

The only way the graphs intersect with a non-negative y value is if the quadratic equation x^2 - 3x - 4 has non-negative y values (since x^2 + Lx is always non-negative for non-negative x values). This occurs when x is greater than or equal to zero. Solving for when y is non-negative:

x^2 - 3x - 4 >= 0

The parabola opens upward and has roots at x = -1 and x = 4. Therefore, it will have non-negative y for x >= 4. In this range, the value of L must be such that Lx adds to the y-value of the first quadratic without making it negative.

For x >= 4, Lx is always positive for any L. Therefore, there are infinite values for L making the intersection at non-negative y possible, since L can be any integer.