Question

A straight line has gradient m and passes through the points (0,-2). Find the two values for m for which the line is tangent to the curve y=x^2-2x 7. And for each value of m, find the coordinates of the points where the line touches the curve

Answer 1

m=8 m=12

Explanation:

1. y=x^2-2x +7
2. dy/dx=2x-2
3. 2(x-1)
4. same as (x-1)(x+1)
5. meaning x can be 1 and another negative 1
6. Soo substitute for the two values of x in the equation y=x^2-2x+7
7. hence coordinates are (1,6) and (-1,10
8. Soo the two m are

• (1,6)(0,-2) change in y/change in x

1. -2-6/0-1
2. -8/-1
3. m =8

• (-1,10)(0,-2) change in y/change in x

1. -2-10/0-1
2. -12/-1
3. m=12