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

Hi,

Explanation:

The curve is y=x^²-2x+7

The tangent has for equation: y=mx-2

Let's find the intersection of the tangent and the curve.

`mx-2=x^2-2x+7\\\\x^2+(-2-m)x+9=0\\\Delta=(-2-m)^2-4*9=m^2+4m-32\\The\ point\ must\ be \ unique: \Delta=0\\\\m^2+4m-32=0\\\delta=4^2+4*32=144=12^2\\\\m=(-4-12)/(2) =-8\ or\ m=(-4+12)/(2) =4\\\\x^2+(-2-m)x+9=0\ and\ \Delta=0 = > x=(2+m)/(2) \\\\if\ m=-8\ then\ x=(2+(-8))/(2)=-3\ and\ y=mx-2=(-8)*(-3)-2=22\\\\if\ m=4\ then\ x=(2+4)/(2)=3\ and\ y=mx-2=4*3-2=10\\\\\\Coordinates\ of\ the\ points\ are (-3,22)\ and\ (3,10)\\`

Answer 2

when m=2 or -2, the line is tangent to the curve at the points (2.7) and (0,7)

Explanation: