Consider the function f(x) = −x 2 + 4x − 2.

Question

a)Sketch the graph of f in the interval [−1, 5]. (6 marks)

b) Find the equation of the tangent line to the graph of f at x = 3, and show it on the graph of f. (4 marks)

Answer 1

f(x) = -x^2 +4x -2

=-(x^2 -4x +4) -2 +4

=-(x-2)^2 +2

is a downward opening parabola with vertex = (2,2) = maximum point

axis of symmetry is x=2.

y intercept =-2 or (0,-2)

x intercepts are found by setting f(x)= 0 and solving for x

x^2-4x+2 =0

(x-2)^2 =-2+4 =2

x-2 =sqr2

x=2+sqr2 or 2-sqr2

x intercepts are about 3.14 and 1.59 or (1.59,0) and (3.14,0)

f(-1) = -1-4-2 =-7

f(5) = -25+20-2 = -7

those are the points (-1,-7) and (5,-7) points symmetric about x=2

plot those points, the intercepts and vertex, and draw a smooth parabolic curve through them.

tangent line at x=3

will have slope = f'(3) = -2x +4 = -2(3)+4 =-2

and go through the point with y coordinate f(3) =-(3)^2 +4(3) -2 = -9+12-2 =1

and x coordinate 3, the point (3,1)

y=-2x + b, plug in the point (3,1) to solve for b

1=-2(3) + b

b =1+6 =7

y=-2x+7 or

2x+y = 7 is the tangent line to the parabola when x=3