Find the tangent and normal to the curve y=(4-x)(x+2) at the point (2, 8)

-Tangent is a straight line that touches, but does not intersect, the curve at the point (2,8).

We need to find the gradient of the curve at the point (2, 8). To do this, expand the equation, differentiate it and then sub in the point (2, 8) to find the gradient.

y=-x^2 + 2x + 8                         

dy/dx= -2x + 2                           

At (2,8) dy/dx = -2. Therefore the gradient of your tangent is -2.

To find the tangent equation, use the formula y= mx+c with m=-2, x=2, y=8. Subbing in these values gives c=12 and therefore the tangent equation is y=-2x + 12.

Normal to the curve is a line that cuts the curve at 90 degrees. This is also known as perpendicular.

The gradient of a perpendicular line is -1/m.

Therefore the gradient is -1/(-2)=1/2 We now use y=mx+c with m=1/2, x=2, y=8 to find the normal equation. This gives c=7 and therefore the normal equation is y=x/2 + 7.