Let $f$ and $g$ be the functions defined by $f(x)=x+2$ and $g(x)=(x^2)−a$, where $a$ is a positive constant. What are all values of $a$ for which the graphs of $f$ and $g$ have exactly one point of intersection?
The solution for this is supposed to be $a > 4$.
I've been agonizing over proving this for days.
I'm positive the solution can only be $-(9/4)$.
Here is my proof:
$$ x - 2 = x^2 - a \implies 0 = x^2 - x - 2 - a$$
The discriminant must be zero for this to only have one solution:
$$\begin{align} b^2 - 4ac = 0 &\implies (1^2)-4(1)(-2-a) = 0 \\ &\implies 1 + 8 + 4a = 0 \\ &\implies 9 + 4a = 0 \\ &\implies a = (-9/4) \end{align}$$
If $a$ is a positive constant then the parabola of $x^2$ will be shifted down the Cartesian plane, while the linear $x+2$ will remain above the origin, meaning that a positive $a$ will always result in two solutions in this system.
Why is the correct answer $a > 4$?
So, we are investigating the points of intersection of $f(x) = x+2$ and $g(x) = x^2 -a$ for constant $a$. These can be analogized as the roots to
$$g(x) - f(x) = x^2 - a - x - 2 = x^2 - x + (-a-2) \tag 1$$
To find the roots, we use the quadratic formula here:
$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-a-2)}}{2(1)} = \frac{1\pm\sqrt{1+4(a+2)}}{2} = \frac{1 \pm \sqrt{4a+9}}{2}$$
When do quadratics have double roots? That is when their discriminant is zero. A double root in this scenario also means that $g(x) - f(x)$ is zero at exactly one point $x$.
Here, clearly, the $a$ that ensures a zero discriminant is $a = -9/4$.
So, in short, your proof is correct.
To further emphasize this, notice the graphs below:
The linear equation is $f$, the blue parabola is $g$ for $a=-9/4$, and the green for $g$ when $a = 5 > 4$.
I'm honestly not sure what could be drawn from $a>4$ or what might have been intended. But regardless, to the question as posed, $a=-9/4$ is correct, and $a>4$ is a mistake.