The function f is defined by f(x)= sqrt (25-x^2) for -5 less than/equal to x less than/equal to 5
a. write an equation for the line tangent to the graph of f at x=-3

Question

The function f is defined by f(x)= sqrt (25-x^2) for -5 less than/equal to x less than/equal to 5
a. write an equation for the line tangent to the graph of f at x=-3

anonymous · Answer

take the derivative, replace $x$ bu $-3$

anonymous · Answer

$$f(x)=\sqrt{25-x^2}$$
$$f'(x)=\frac{-x}{\sqrt{25-x^2}}$$

anonymous · Answer

i am going to guess you get $\frac{3}{4}$ but you can check it

anonymous · Answer

what i mean is that the slope will be $\frac{3}{4}$

anonymous · Answer

it is clear that this function is the upper half of the circle with radius 5 centered at the origin?

anonymous · Answer

It's not specified in prompt if that's what you mean, but i guess it would be