# Gradient of the point with x coordinate

I am not even sure where to begin on solving this question

f(x) = $ax^2$ + bx + c, where a = -32, b = 9 and c = 14. What is the gradient of the point, with x coordinate 11, on the graph of y = f(x)?

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Do you know about differentiation? – user39572 Oct 17 '12 at 14:26
@J.G. Yes I know a little about differentiation,This is more of a fixed point topic – JackyBoi Oct 17 '12 at 14:28
Then differentiate $f$, the gradient will be $f'(11)$ (given any differentiable function $f$, the gradient at $x=a$ is defined as $f'(a)$) – user39572 Oct 17 '12 at 14:29

Hint: plug the given $a,b,c$ into your function, differentiate with respect to $x$, and set $x=11$

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Would my answer be correct? – JackyBoi Oct 17 '12 at 14:41
@JackyBoi: Yes it would. Doing the differentiation symbolically saved you some computation, usually a good thing. – Ross Millikan Oct 17 '12 at 14:43

Ya actually the answer was something like this according to the given formula

In general, if f(x)= $ax^2$ + bx + c then the gradient at any point on the graph of y=f(x) is given by f prime (x) = 2ax + b

so in the above sums case it will be 2(-32)(11)+9 = -695

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