Finding the gradient of the tangent to a parabola how to use the quadratic theory to find the gradient of the tangent to:
y=x^2 at the point (2,4).
how to go about solving this simple problem as I am mixed up. 
Thank you.
 A: Hints:
(1) The slope to the tangent line to the graph 0f $\;f(x)\;$ at a point $\;(a, f(a)\;$ on 
the graph is given, assuming $\;f\;$ is differentiable at that point, by $\;f'(a)\;$
(2) For any $\;n\in\Bbb R\;$, we have that $\;(x^n)'=nx^{n-1}\;$
A: Note that in modern analysis one usually defines the line tangent to $f$ in $x_0$ as $$y=f'(x_0)(x-x_0)+f(x_0).$$
So, for the sake of rigour, you should prove that the result obtained with less advanced instruments is equivalent. However, let's define the tangent line as the one which is not vertical and intersects the parabola in one and only one point, $(x_0,ax_0^2)$. We have two equations for the tangent: $$y=m(x-x_0)+ax_0^2,$$
that is, the line passes through $(x_0,ax_0^2)$. The second one is obtained by considering the quadratic equation $$ax^2=m(x-x_0)+ax_0^2,$$
which is satisfied in the intersections of the parabola and the line. The second contraint is given by the condition $\Delta(m)=0$, where $\Delta$ is the discriminant of this quadratic (and it depends on the parameter $m$, having fixed $a$).
A: The equation of a line is $f(x)=mx+b$ where $m$ is the slope.
The equation of a tangent line to a function $f(x)$ on $a$ is $f(x)=f'(a) (x-a) + f(a) = f'(a)\ x+[f'(a)\ a+f(a)]$, which has the form $y=mx+b$. The slope is $f'(a)$
$$f'(x)=\frac{d}{dx}(x^2)=2x$$
You have $a=2$, becuase the point is $(a,f(a))=(2,4)$
So, the slope $m$ is $$m=f'(a)=f'(2)=2(2)=4$$
A: As soon as you ask for the gradient of a tangent to a curve, you are getting into the territory of calculus. But don't worry, it's not too hard.  
We start by finding the gradient of a line through the two points $A$ and $B$, where $A=(2,4)$ and $B$ is any nearby point on the curve. So take some small number $h \ne 0$, let $x_B = 2 + h$, and put $y_B = x^2 = (2+h)^2$. Now we have a point $B=(x_B,y_B)$ that is on the curve, and close to the point $A=(2,4)$. (Here you might like to draw yourself a picture.)  
Now we calculate the gradient of the line through $A$ and $B$: this is just
$$(y_B-y_A)/(x_B-x_A) = ((2+h)^2-4)/((2+h)-2) = (h^2+4h)/h = h + 4$$
Now you can see that as the point $B$ gets closer and closer to $A$ (or in other words, as $h$ gets closer and closer to $0$), this gradient gets closer and closer to $4$. In fact we can make it as close to $4$ as we like by making $h$ small enough.
So the gradient of the curve at $A$ is $4$. And that's calculus!
