Precalc Question - Function Values?
Hello! Will you please help me with this question? For context, I am taking Dr. James Murphy's Modern States Precalculus course to practice for the CLEP exam.
- Anonymous2 months agoFavorite Answer
This particular question is worded incorrectly for the given solutions.
You are given f(x) = x + 2 and g(x) = x^2 - a. Find the point(s) of intersection.
x^2 - a = x + 2
x^2 - x - a - 2 = 0
For any parabola and a linear function, there will be either no intersections, one point of tangency, or two intersections. Recall the discriminant of a quadratic equation. If the discriminant is negative, there are no real solutions, and the zeroes are complex conjugates. If the discriminant is zero, there is one real, repeated root (the vertex of the parabola is on the x-axis). If the discriminant is positive, there are two real roots. If instead of intersecting the line y=0 (the x-axis), the parabola is now intersecting the line y = x+2. The discriminant is equal to 0 when there is exactly one point of intersection, when the line is tangent to the parabola (may or may not be at the vertex).
0 = (-1)^2 - 4(1)(-2-a)
0 = 1 - 4(-2-a)
0 = 1 + 8 + 4a
0 = 4a + 9
4a = -9
a = -9/4
I am guessing the question was supposed to be asking,
Given f(x) = x + 2 and g(x) = x^2 - a. For what values of a > 0 is there exactly one point of intersection ABOVE THE X-AXIS.
Since f(x) = x + 2 is positive when x > -2
For g(x), the vertical shift of a units moves the vertex along the y-axis
We are asked to find the values of a that result in one point of intersection above the x-axis and one below the x-axis. This is the same as finding when one zero of g(x) is less than -2.
g(-2) = 0
(-2)^2 - a = 0
4 - a = 0
If a=4, -2 is a root of g(x) (the other root is 2). If a > 4, the negative root of g(x) is less than -2, and therefore the intersection with f(x) will occur where f(x) < 0 and g(x) < 0. Then there is exactly one point of intersection BELOW the x-axis and one point of intersection ABOVE the x-axis.