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 $h+9$ ?

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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 $h+9$ ?

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Answer 1 · 2017-07-15T21:30:34

We have:

$y=x^2+3x$

When $x=3$ we have:

$y =3^2+3(3)$
$\ \ =9+9$
$\ \ =18$

When $x=3+h$ we have:

$y = (3+h)^2+3(3+h)$
$\ \ = 9+6h+h^2 + 9 + 3h$
$\ \ = h^2+9h+18$

So the gradient of the secant line is given by:

$m_(sec) = (Delta y)/(Delta x)$

$" " = (y_(3+h) - y_h)/((3+h)-(3))$

$" " = ((h^2+9h+18) - (18))/(3+h-3)$

$" " = (h^2+9h)/(h)$

$" " = h+9 \ \ \$ QED

Conclusion

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

$m_(tan) = lim_(h rarr 0) m_(sec)$

$" " = lim_(h rarr 0) (h+9)$

$" " = 9$

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:

$y=x^2+3x => dy/dx = 2x + 3$

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

$[dy/dx]_(x=3) = 2(3) + 3 = 9$

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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 $h+9$ ?

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

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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 $h+9$ ?