Three variable systems if equations.

Question

Given the quadratic function $y=x^2 + 4$ and the linear function $y=x + b$, determine all the possible values of $b$ that would result in a system if equations with two solutions, exactly one solution, and no solution. How would one accomplish this?

Answer 1

Substitute y = x + b into y = x^2 + 4 to get x + b = x^2 +4. This is a quadratic x^2 - x - b + 4 = 0. What tells you about the number of real solutions of a quadratic.

Answer 2

minimum b that you can find is when curve and line are tangent. so find min b by solving the equation " curve=line " , discriminant is equal or greater than zero .
x^{2}+4=x+b rightarrow x^{2}-x+4-b=0 rightarrow Delta ge0 rightarrow 1-4(4-b)ge0 rightarrow b ge \frac{ 15 }{4 }