What is the positive value of k for which the graph is $y=x^2 -2kx+16$ is tangent to the $x$-axis?
My approach to solving this is that since the parabola is tangent to the $x$-axis, I can find the value of $x$ by using $-\frac{b}{2a}$. Once I have $x$ I can solve for $k$. Is this the right approach?
If the parabola is tangent to $x$-axis it means $y=0$ so you must solve \begin{align*} 0 &= x^2-2kx+16 \end{align*} for $x$ and $k$. It also means there is only one value possible for $x$, otherwise the parabola will cross the $x$ axis instead.
At this point you can choose (a) applying the formula for quadratic equations, or (b) factor the expression. We'll do both so you can choose the one you like the most.
(a) applying the formula for quadratic equations requires $a=1$, $b=-2k$ and $c=16$ \begin{align*} x &= \frac{ -(-2k) \pm\sqrt{ (-2k)^2 -4(1)(16) } }{ 2(1) }\\ &= \frac{ 2k\pm\sqrt{4k^2-4\cdot 16} }{ 2 }\\ &= \frac{ 2k \pm 2 \sqrt{k^2-16} }{ 2 } \\ &= k \pm \sqrt{k^2 -16} \end{align*} You need an unique value for $x$ so $k^2-16$ must be zero, therefore $k=4$ and $x=4$.
(b) to factor the expression, \begin{align*} x^2-2kx+k^2 &= x^2-2kx+16 \\ (x-k)^2 &= x^2-2kx+16 \end{align*} means that $k=4$ and solving the expression $(x-4)^2=0$ gives $x=4$.
The parabola is tangent to the $x$-axis if and only if the equation $x^2-2kx+16=0$ has a double root, i.e. if its (reduced) discriminant $\Delta'=k^2-16$ is equal to $0$.
The parabola being tangent to the x-axis means that the equation just have one root. This happens when $\Delta = b² -4ac$ is zero. If you calculate $\Delta$ with the information given, you can find $k$ so.
In order for the Parabola $y=x^2 - 2kx + 16$ to be tangent to the $x$-axis, then its equation must have only one root (one "common" point with the axis).
Therefore, its discriminant must be equal to $0$ :
$$\Delta = 4k^2 - 4\cdot 16=0 \Leftrightarrow 4k^2 - 64 = 0 \Leftrightarrow k = +- 4 $$
The positive value of $k$ that you need is : $k=4$.
If $y = x^2 -2kx + 16$ is tangent to the $x$-axis then there's a double root at $y=0$.
Therefore $x^2 -2kx + 16$ is a perfect square, i.e., it can be factorised as
$$x^2 -2kx + 16 = (x - u)^2$$
Thus
$$x^2 -2kx + 16 = x^2 - 2ux + u^2$$
for all $x$. Equating coefficients,
$$k = u \text{ and } 16 = u^2$$
Since we only want the positive $k$ we get
$$k = u = 4$$
