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Thank you for taking my question. I am home schooling my son in Calculus, but for the first time, I am having trouble with today's lesson material; I am afraid that I can't teach it properly if I haven't grasped it 100%. I was hoping you could show the answer to the following question and include your steps/processes a bit.

It is:

f(x)=-x^(2)+4

Let f(x) be the function given by f(x)=-x^(2)+4 The graph crosses the y-axis at P(0,4) and the x-axis at Q(2,0). What is an equation for the tangent of the graph of f at point Q (slope-intercept form).

Thank you for your help,

John

Hi John,

The slope of the function at a point (which is also the slope of the tangent line there) is given by the value of the derivative at that point. For the function f(x) = -x² + 4, the derivative is given by

f'(x) = -2x

and so at the point (2, 0), the slope is

f'(2) = -2(2) = -4

Now, in coordinate geometry the equation of a line with slope m passing through a point (x1, y1) is given by

m = (y - y1) / (x - x1)

So, the equation for the tangent at the point Q is

-4 = (y - 0) / (x - 2)

-4 = y / (x - 2)

y = -4(x - 2)

y = 8 - 4x

If you need anything else do let me know. Goodluck.

Regards

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