IB Maths Studies Equation of Tangent

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IB Maths Studies Equation of Tangent

IB Mathematical Studies SL – Calculus, Derivatives, Differentiation, Equation of Tangent

How can we find the equation of the tangent to the curve $f(x)=3x^2-5x$ at the point $(1,-2)$.

Thanks
hill

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Joined: Tue Mar 19, 2013 8:49 pm

Re: IB Maths Studies Equation of Tangent

IB Mathematical Studies SL – Calculus, Derivatives, Differentiation, Equation of Tangent

We know that the gradient of a curve $f(x)$ at any point $(x_{0},y_{0})$ is equal to the gradient of the tangent ($m_{T}$) to the curve at that point.

So about your question we have that
$f'(x)=6x-5$

and the gradient of the tangent ($m_{T}$) at the point $(1,-2)$ is

$m_{T}=f'(1)=6-5=1$

and the equation of the tangent is given by the following formula

$y-y_{0}=f'(x_{0})(x-x_{0})$

where in your case

$(x_{0},y_{0})=(1,-2)$

and finally the equation of the tangent is

$y-(-2)=1(x-1) \Rightarrow y=x-3$

Hope these help!!
elizabeth

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Joined: Mon Jan 28, 2013 8:09 pm

Re: IB Maths Studies Equation of Tangent

Thanks Elisabeth!!
hill

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Joined: Tue Mar 19, 2013 8:49 pm