Question

How do you find the tangent line of `f(x) = 3-2x^2` at x=-1?

Answer 1

Take the derivative and plug in the desired `x=-1` to get the slope. Then back-solve for the y-intercept. You will find the expression is:

Explanation:

The slope-intercept equation for a given line is `y=mx+b`. We know by definition that a tangent line has the same value as the original function at the point of tangency, and the slope is the same as the functions' at that point.

We can calculate the slope from the derivative of the function:

`f(x)=3-2x^2`

`(dy)/(dx)f(x)=f'(x)=-4x`

`f'(-1)=-4(-1)=color(blue)(4)`

Now that we have our slope, we can evaluate the original function and solve for the y-intercept:

`f(-1)=3-2(-1)^2`

`f(-1)=3-2`

`f(-1)=1`

`f(-1)=4(-1)+b`

`1=-4+b`

`color(blue)(b=5)`

Now, we know the slope and intercept, and can write the equation for the tangent line:

`color(green)(y=4x+5)`