How to find the line tangent to two graphs

[OmniXN]

The graphs of the quadratic functions

f(x) = 6-10x^2

and

g(x)=8-(x-2)^2

Find the lines simultaneously tangent to both graphs.
(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation: (Two decimal places of accuracy.)
(b) The other line simultaneously tangent to both graphs has equationTwo decimal places of accuracy.)

I know I need to find some version of y=m(x-a)+b, But I'm not sure how to go about it.

Thank you for any help provided.

 

[Ishuda]

The graphs of the quadratic functions

f(x) = 6-10x^2

and

g(x)=8-(x-2)^2

Find the lines simultaneously tangent to both graphs.
(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation: (Two decimal places of accuracy.)
(b) The other line simultaneously tangent to both graphs has equationTwo decimal places of accuracy.)

I know I need to find some version of y=m(x-a)+b, But I'm not sure how to go about it.

Thank you for any help provided.

What is m, a, and b? The equation for a line tangent to a curve at the point (x₀, f(x₀)) [(x₁, g(x₁))] is
y_f(x) = f(x₀) + f'(x₀) (x - x₀)
and
y_g(x) = g(x₁) + g'(x₁) (x - x₁)
Now for a line to be simultaneously tangent to both line, that means (x₀, f(x₀)) and (x₁, g(x₁)) must be on the same line which means they must both have the same slope or f'(x₀) = g'(x₁). That will give you an equation for x₀ in terms of x₁ (or vice versa) and since they are the same line, f(x₀) - f'(x₀) x₀ = g(x₁) - g'(x₁) x₁. Use the fact that the slopes are equal and the value of x₀ in terms of x₁ to evaluate f(x₀) and f'(x₀) (or vice versa) to get a quadratic equation which you can solve for the two roots.

But sure seems like there should be some easier way. Anyone?