# Show that the gradient of the secant line to the curve y=x^2+3x at the points on the curve where x=3 and x=3+h is h+9?

Jul 15, 2017

We have:

$y = {x}^{2} + 3 x$

When $x = 3$ we have:

$y = {3}^{2} + 3 \left(3\right)$
$\setminus \setminus = 9 + 9$
$\setminus \setminus = 18$

When $x = 3 + h$ we have:

$y = {\left(3 + h\right)}^{2} + 3 \left(3 + h\right)$
$\setminus \setminus = 9 + 6 h + {h}^{2} + 9 + 3 h$
$\setminus \setminus = {h}^{2} + 9 h + 18$

So the gradient of the secant line is given by:

${m}_{\sec} = \frac{\Delta y}{\Delta x}$

$\text{ } = \frac{{y}_{3 + h} - {y}_{h}}{\left(3 + h\right) - \left(3\right)}$

$\text{ } = \frac{\left({h}^{2} + 9 h + 18\right) - \left(18\right)}{3 + h - 3}$

$\text{ } = \frac{{h}^{2} + 9 h}{h}$

$\text{ } = h + 9 \setminus \setminus \setminus$ QED

Conclusion

Note that as $h \rightarrow 0$ then the secant line becomes the tangent, so in the limit we have:

${m}_{\tan} = {\lim}_{h \rightarrow 0} {m}_{\sec}$

$\text{ } = {\lim}_{h \rightarrow 0} \left(h + 9\right)$

$\text{ } = 9$

It should be clear that this final result comes as a direct result of the definition of the derivative. We can also use our knowledge of Calculus to validate this, as:

$y = {x}^{2} + 3 x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 3$

And so, when $x = 3$, we have:

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = 3} = 2 \left(3\right) + 3 = 9$