Question

For what values of a and b will the parabola y=x^2 + ax + b be tangent to the curve y=x^3 at point (1,1)?

Answer 1

The values are `a=1," "b="–"1`.

Explanation:

To solve for the two variables `a` and `b`, we require a system of two equations. (Remember those? Well, they're coming in handy here!)

In order for two curves to be tangent to each other at a given point, they must have the same tangent line at that point. (If their tangent lines at the given point were different, the two curves would "pass through" each other rather than just "graze" each other.)

The slope of the tangent line to `y=x^3` at the point `(x_0, y_0)=(1,1)` is the first derivative of `y` (that is, `y'`) at `x_0=1`:

`y'=3x_0^2=3(1)^2=3`

Thus, the slope-point form of the tangent line required is:

`y-y_0=m(x-x_0)`
`y-1=3(x-1)`

For `y=x^2+ax+b` to share a tangent line at the point `(1,1)`, all we need is for the two slopes to be identical. (That way, the two tangent lines will have the same slope-point form).

Thus, we need `y=x^2+ax+b` to satisfy two conditions:

1. It must pass through `(x_0,y_0)=(1,1).`
2. It must have a slope (`y'`) of `3` at this point.

These will give us our two equations in two unknowns. The first one gives us

`color(white)=>y_0=x_0^2+ax_0+b`
`=>1=(1)^2+a(1)+b`
`=>0=a+b`

and the second one gives us

`color(white)=>y'=2x_0+a`
`=>3"  "=2(1)+a`
`=>a=1`

Looks like we know exactly what `a` is. All we need is to plug this value into the first equation and we'll find `b`:

`a+b=0" "<=>" "b="–"a`
`"                      "=>"    "b="–"1`

Here is a graph of the two curves:
graph{(y-x^3)(y-x^2-x+1)=0 [-4.937, 4.93, -1.716, 3.22]}

Bonus:

The quick way to solve this problem is to realize that both derivatives must be equal at `(1,1)`, so we equate the two as follows:

`color(white)=>3x^2=y'=2x+a`

`=>3(1)^2=2(1)+a`

`=>3-2=a" "=>" "a=1`

Then, we just have to plug in the known values for `x, y`, and `a` into `y=x^2+ax+b` to find `b`:

`color(white)=>y=x^2+ax+b`

`=>1=(1)^2+(1)(1)+b`

`=>1-2=b" "=>" "b="–"1`.