PreCalculus
posted by Shirley .
Im having trouble with this.
Find an equation of the tangent line to the parabola at the given point and find the xintercept of the line.
x^2=2y (4,8)

y = (1/2) x^2
The slope of the curve is dy/dx = x
At x = 4, the slope is 4.
For a line though (4,8) with slope = 4, the equation is
(y8) = 4(x4)
Rewrite that in standard y = mx + b form. 
well, from the above equation y = (x^2)/2
to get an equation of a line you need a slope and a point. You have a point and to get the slope we need to get dy/dx
and sub in the value for x in the point given [in this example 4]
Then sub your point (x1,y1) and the slope (m) into
y  y1 = m(x  x1)
and then just tidy up
Hope that helps