Finding the values of k for an equation which is a tangent to a curve

Question

I have been faced with the following question:

Find the values of $k$ for which $y = kx - 2$ is a tangent to the curve $y = x^2 - 8x + 7$

I managed to figure this out by treating them as simultaneous equations and then calculating the discriminant so that k either has the value $-14$ or $-2$ however I was wondering if there was another possible method to answer this perhaps by way of differentiation.

EDIT

And using differentiation how is it possible to get the answer that k is either $-14$ or $-2$?

Answer 1

Hint: The tangent at $(a,b)$ is $y=(2a-8)(x-a)+b$.

Since $(a,b)$ is a point on the parabola, we have $b=a^2-8a+7$. Hence the tangent equation is $y=(2a-8)x-a^2+7$, i.e. $k=2a-8$ and $a^2-7=2$.

Answer 2

HINT

For the graphs $y=f(x)$ and $y=g(x)$ to be tangent at $x=a$ you need

$$f(a)=g(a)\text{ and }g'(a)=f'(a).$$

Answer 3

Find the values of $k$ for which $y = kx - 2$ is a tangent to the curve $y = x^2 - 8x + 7$

• Where do the equations intersect? Solve for $kx - 2 = x^2 - 8x + 7$.

• Where is the slope of the line $y = kx - 2$ $\Big($ Slope = $ \frac{d}{dx}(kx - 2) = k\Big)\;$ equal to $\;\frac{d}{dx}(x^2 - 8x + 7) = 2x - 8\;$?