A straight line passes through the point (2,0) and has gradient m. Write down the equation of the line.

Question

Find the two values of m for which the line is a tangent to the curve y=x^2-4x+5. for each value of m, find the coordinates of the point where the line touches the curve.

Answer 1

x^2 -4x +5 is a parabola with slope = m =y- = derivative of x^2 -4x +5 = 2x-4

a point of tangency would be (2x,-4, y)

a line through (2,0) with slope m

would be 0=2m +b

b=-2m

y =mx -2m where m=2x-4 with 2nd x from the parabola, the x coordinate of the tangency point

it helps to draw a rough sketch of the point, line and parabola

at (0,5) the parabola has slope = -4

a line through (0,5) and (2,0) has slope -2.5

at (1,2) the parabola has slope = -2

a line through (1,2) and (2,0) has slope = -2

m=-2 and +2

the points of tangency are (1,2) and (3,2)

tangency lines are through the point (1,2) and (2,0)

and through (3,2) and (2,0)

equations of the tangency lines are

y =-2x+4 and y =2x-4

y=x^2-4x+5 has a y intercept = 5 or the point (0,5), but no x intercepts. the entire parabola is everywhere above the x axis

minimum point or vertex is (h,k)= (2,1), where (x-h)^2 + k = (x-2)^2 +1= x^2 -4x +4+1= x^2-4x+5

(2,0) is directly below the vertex, on the x axis

the two tangent lines will have the same absolute values, but opposite signs, as one goes uphill and the other downhill