## Derivatives equation tangent IB math HL

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### Derivatives equation tangent IB math HL

Differentiation, Derivatives, gradient, tangent - IB Mathematics HL

How can we find the equation of the tangent to the curve with equation

$f(x)=x^4+5x^2+3x+4$ at the point $(1,13)$?

Thanks
Jack

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

### Re: Derivatives equation tangent IB math HL

Derivatives equation tangent IB math HL - IB mathematics Higher Level

$f(x)= x^4+5x^2+3x+4\Rightarrow$

$f’(x)=4x^3+10x+3 \Rightarrow$

The gradient of the curve at $(1,13)$ is

$f’(1)= 4(1)^3+10(1)+3=17$

The equation of the tangent on this point is given by the equation

$y-y_{0}=m_{t}(x-x_{0})$

where $m_{t}=17$ is the gradient of the tangent

and the touch point is $(x_{0}, y_{0})=(1,13)$.

So, the equation of the tangent at the given point is

$y-13=17(x-1) \Rightarrow y=17x-4$

Hope these help!!
elizabeth

Posts: 0
Joined: Mon Jan 28, 2013 8:09 pm

### Re: Derivatives equation tangent IB math HL

Thank you!
Jack

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