Parabola with given line. Please help!

Question

Given the parabola y = −(x − 2)^2 and the line: y = x − 8 .

a. Graph them on the same set of axes and find the parabola's negativity interval.

b. Find the interval where the graph of the parabola is above the graph of the line.

c. Find the equation of a tangent line to the graph of the parabola, which has a slope of 4.

Answer 1

y=-(x-2)^2 and y =x-8

find where the parabola is below the x axis, where y values are negative

vertex = maximum point = (2,0), the parabola is always below the x axis except at the vertex

"negativity" of the parabola is in the intervals (-infinity,2)U(2,+infinity)

it's above the line x-8 when

-(x-2)^2>x-8

-x^2 +4x -4 > x-8

x^2 -3x -4 >0

end points of the interval are the x values of the above inequality, -1 and 4

(x-4)(x+1) >0

interval is (-1,4) or -1<x<4

the parabola is y=-(x-2)^2 = -(x^2-4x+4)= -x^2 +4x +4

the parabola has a slope = 4 when y'=-2x+4 = -4, when x=0 and y =-4, the point (0,-4)

the tangent line is

4=(y+4)/x

or

y=4x-4

y=-4