There are a few ways in which you can do that and I will show you two of them.

**I. method**

First calculate value of the function at point `x_0`:

`f(4)=6`, `f(9)=91` (this is `y_0` in next formula (1))

Now calculate derivative of function `f`:

`f'(x)=4x-9`

Now calculate vlue of derivative at `x_0`:

`f'(4)=7`, `f'(9)=27`

...

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There are a few ways in which you can do that and I will show you two of them.

**I. method**

First calculate value of the function at point `x_0`:

`f(4)=6`, `f(9)=91` (this is `y_0` in next formula (1))

Now calculate derivative of function `f`:

`f'(x)=4x-9`

Now calculate vlue of derivative at `x_0`:

`f'(4)=7`, `f'(9)=27`

Now use the following formula (equation of tangent through a point on a graph) to calculate your approximation:

`y-y_0=f'(x)(x-x_0)` **(1)**

For `x_0=4`

`y-6=7(x-4)=>y=7x-22`

For `x_0=9`

`y-91=27(x-9)=>y=27x-152`

**II. method**

Here we will use Taylor series. This is a more general method because in this way we can calculate polynomial of `n`-th degree that approximates our function at a given point.

Taylor formula:

`y=f(x_0)+(f'(x_0))/(1!)(x-x_0)+(f''(x_0))/(2!)(x-x_0)^2+(f^((3))(x_0))/(3!)(x-x_0)^3+cdots `

Because we are looking for linear approximation we will use only first two ellements of above sum.

For `x_0=4`

`y=6-7(x-4)=7x-22`

For `x_0=9`

`y=91-27(x-9)=27x-152`

As you can see result of both methods is the same because in both cases we use tangent to approximate our function. But we didn't have to use tangent. Sometimes it's better to use secant line to approximate function on a given interval.