Looking for a quicker way to solve a quadratic problem

Question

Let $f(x) = x^2 – 6x + 3$ and $g(x) = k – f(1 – x)$ for some constant $k$. It is known that $y = g(x)$ touches x-axis at only one point $B = (b, 0)$.

(i) Find the values of $b$ and $k$. (ii) Solve $2f(x) + g(x) – 2 = 0$.

My attempt:-

$g(x) = k – f(1 – x) =$ … after expansion… $= –x ^2 – 4x + 2 + k$

“$y = g(x)$ touches x-axis at only one point $B = (b, 0)$” implies $b = \dfrac{-(-4)}{2(-1)} = -2$.

That statement also implies $g(x) = –x ^2 – 4x + 2 + k = 0$ has equal roots. Then, $k = … = –6$.

My questions are:-

1) Can I get the same result without expanding the $(1 – x)^2$ term?

2) It seems the equation in (ii) is not well related to (i), so what is the quickest way to do (ii)?

Answer 1

1) Notice that $f(x)$ is a parabola with vertex at $x=3$. The graph of $f(1-x)$ is the graph of $f(x)$ shifted to the left by 1, and then reflected over the $y$-axis. So it's a parabola with vertex at $x=-2$. So you can see right away that $b=-2$, since the only way it can have one root is if it just touches the $x$-axis at the vertex of the parabola.

2) Both $f$ and $g$ are quadratic, so I would just expand it out and use the quadratic formula.

Answer 2

Here's a way to speed up part (i), without expanding $(1-x)^2$.

$g$ only touches the $x$ axis in one spot ($x=b$), so its stationary point (zero of its derivative) is that spot. But by chain rule: $$ g'(x) = f'(1-x)=-2x-4 $$ Thus, $$ g'(b)=-2b-4=0 \;\;\;\implies\;\;\; b=-2 $$ For (ii), I don't see an obvious speed up.