## IB Maths SL, Equation of Tangent

Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths SL Revision Notes

### IB Maths SL, Equation of Tangent

IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

How can we find the equation of the tangent to the curve $f(x)=x^2-5x$ at the point $(1,-4)$.

Thanks
Oliver

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Joined: Mon Jan 28, 2013 8:08 pm

### Re: IB Maths SL, Equation of Tangent

IB Mathematics SL – Derivatives, Differentiation, Tangent and Normals

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.

$f'(x)=2x-5$

and the gradient of the tangent ($m_{T}$) at the point $(1,-4)$ is

$m_{T}=f'(1)=2-5=-3$

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

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

$(x_{0},y_{0})=(1,-4)$

and finally the equation of the tangent is

$y-(-4)=-3(x-1) \Rightarrow y=-3x-1$

Hope these help!!
miranda

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Joined: Mon Jan 28, 2013 8:03 pm

### Re: IB Maths SL, Equation of Tangent

thanks miranda!!
Oliver

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm