Question

Let f(x) = 2(x- h1)*2 + k1 and g(x) = ( x- h2)*2 + k2, where h1, h2,k1,k2. The vertex of the graph of f is at (m,-m*2) and the vertex of the graph of g is at (-m,-m), where 0 < m < 1. The graphs of f and g intersect at exactly one point. Find the value of m.

Answer 1

To find the value of m, we need to first find the equations of f(x) and g(x) using the information given.

From the vertex of f, we know that (h1, k1) = (m, -m*2), so the equation of f(x) can be written as:

f(x) = 2(x - m)^2 - 2m

From the vertex of g, we know that (h2, k2) = (-m, -m), so the equation of g(x) can be written as:

g(x) = (x + m)^2 - m

We are given that the graphs of f and g intersect at exactly one point, so we can set f(x) = g(x) and solve for x:

2(x - m)^2 - 2m = (x + m)^2 - m

Expanding both sides:

2(x^2 - 2mx + m^2) - 2m = x^2 + 2mx + m^2 - m

2x^2 - 4mx + 2m^2 - 2m = x^2 + 2mx + m^2 - m

Simplifying and rearranging:

x^2 - 6mx + m^2 + m - 2m^2 + 2m = 0

x^2 - 6mx - m^2 = 0

Since we are given that the graphs intersect at exactly one point, the discriminant of this quadratic equation must be equal to zero:

b^2 - 4ac = (-6m)^2 - 4(1)(-m^2) = 0

36m^2 + 4m^2 = 0

40m^2 = 0

m^2 = 0

m = 0

Since the given condition is that 0 < m < 1, the value of m that satisfies all the conditions is m = 0.