Since you will use several values of $x$, it is probably easiest to find a function for the slope, in terms of $x$, and then plug in the various values for $x$. That is, the slope of the secant line $PQ$ is the rise over run (change in $y$ over change in $x$):
$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$
So, $m(x)$ gives the slope for any particular value of $x$. A practical reason to do this is, for example on a TI-83 or TI-84 or something like it, you can now type in that function to $Y_1$ and then go to the Table and plug in the various $x$ values and you get the slopes immediately. And, since it is so fast, you can check this for more values than the question even asks for to get an even better intuition. Or, you could use Wolfram Alpha to accomplish the same thing. For example, type in:
Evaluate (x^2 + x + 4 - 24)/(x - 4) at x = 4.1, 4.01, 4.001, 3.9, 3.99, 3.999
Back to the problem,
$\begin{align*}
m(4.1) =& \frac{4.1^2 + 4.1 + 4 - 24}{4.1 - 4} = \frac{0.91}{0.1} = 9.1 \\
m(4.01) =& \frac{4.01^2 + 4.01 + 4 - 24}{4.01 - 4} = \frac{0.0901}{0.01} = 9.01 \\
m(4.001) =& \frac{4.001^2 + 4.001 + 4 - 24}{4.001 - 4} = \frac{0.009001}{0.001} = 9.001 \\
m(3.9) =& = \frac{3.9^2 + 3.9 + 4 - 24}{3.9 - 4} = \frac{-0.89}{-0.1} = 8.9 \\
m(3.99) =& = \frac{3.99^2 + 3.99 + 4 - 24}{3.99 - 4} = \frac{-0.0899}{-0.01} = 8.99 \\
m(3.999) =& = \frac{3.999^2 + 3.999 + 4 - 24}{3.999 - 4} = \frac{-0.008999}{-0.001} = 8.999
\end{align*}$
From this, we would probably guess that the slope of the tangent line when $x = 4$ is 9. This does not guarantee that we are right, but assuming the function is reasonably well behaved, we could be pretty confident in this guess.
And, once you learn how to calculate derivatives, you will find this is correct as the slope of the tangent line at any $x$ value is the derivative. Since
$$y'(x) = 2x + 1$$
we see that
$$y'(4) = 9$$
Best Answer
You know that the tangent at the curve is given by
$$y_t = f(a)+f'(a)(x-a)$$
The normal would be a line such that
Given that a line that passes through $(X,Y)$ and has slope $m$ is given by
$$y-Y = m(x-X)$$
...and that two lines of slopes $n$ and $p$ are perpendicular if and only if $m\cdot n=-1$
Can you find $y_n$?
Give the equations to