Posted by **Shirley** on Sunday, March 22, 2009 at 6:21pm.

Im having trouble with this.

Find an equation of the tangent line to the parabola at the given point and find the x-intercept of the line.

x^2=2y (4,8)

- Pre-Calculus -
**drwls**, Sunday, March 22, 2009 at 7:34pm
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

(y-8) = 4(x-4)

Rewrite that in standard y = mx + b form.

- Pre-Calculus -
**GanonTEK**, Sunday, March 22, 2009 at 7:35pm
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

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