Question

How do you find the equation of the line tangent to `y = x^(3) + 2x^(2) - 3x + 2` at the point where x = 1?

Answer 1

The tangent line is:

`y = 4x-2`

Explanation:

For `x=1`:

`y= 1+2-3+2=2`

so the curve goes through the point:

`(bar x, bar y) = (1,2)`

The equation of the tangent line is then:

`y= bar y + f'(barx) (x -bar x)`

where:

`f'(x) = dy/dx = 3x^2+4x-3`

and:

`f'(1) = 3+4-3 = 4`

So the tangent line is:

`y= 2+4(x-1)`

or:

`y = 4x-2`

Answer 2

`y=4x-2`

Explanation:

The point on the graph of `y` where the tangent line will intersect will be at `(1,y(1))`. Here, `y(1)=1+2-3+2=2`, so we know that the tangent line will pass through the point `(1,2)`.

Since the only things we need to know to write the equation of the line are a point it travels through and its slope, we just need to know the slope of the tangent line at `x=1`.

Luckily for this, this is exactly what the derivative of the function tells us. The derivative, evaluated at `x=1`, will tell us the slope of the tangent line at `x=1`.

To find the derivative of `y`, we will need to use the power rule. The power rule states that the derivative of `x^n` is just `nx^(n-1)`. That is, multiply the `x` term by its power, and reduce the power by `1`.

Also remember that constants don't really matter when we're finding the derivative of something--they stay when we differentiate a function. So, where `a` is a constant, we can say that the derivative of `ax^n` is just `a(nx^(n-1))`.

So, when we take the derivative of `y`, we should get:

`dy/dx=3x^2+2(2x^1)-3(1x^0)+2(0)x^-1`

Note that `-3x=-3x^1` and `2=2x^0`.

Then:

`dy/dx=3x^2+4x-3`

So the value of the derivative `dy/dx` at `x=1` is just `3+4-3=4`.

So we know the tangent line has slope `m=4` and passes through the point `(x_1,y_1)=(1,2)`.

The equation for a line passing through `(x_1,y_1)` with slope `m` is given by `y-y_1=m(x-x_1)`, so here the tangent line is `y-2=4(x-1)`, which can be simplified as `y=4x-2`.

Graphed are the original function and its tangent line at `x=1`:

graph{(-y+x^3+2x^2-3x+2)(y-4x+2)=0 [-5, 5, -8, 15.26]}