Create a line that is tangent to f(x)=3-5x^2 and goes through point (-1,-2).

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The given point (-1, -2) lies on the graph of the given function:  f(-1) = 3 - 5 = -2.

In this situation, the equation of the tangent line at this point is

y = f'(-1)(x - (-1)) + f(-1).

Because  f'(x) = -10x,  we obtain  f'(-1) = 10  and the equation becomes

y = 10(x + 1) - 2 = 10x + 8.

The graph is attached.

[the math editor is broken, says "f(x)=x^2" for many formulas]

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We are going to make a tangent line of the form

eq. (1): y(x)=m*x +b

Such that y(x) goes through point (-1,-2). Therefore y(-1)=-2 must be satisfied.

Here m is the slope that is tangent at point (-1,-2) and b is the y-intercept.

First lets find m. 

We need to find the derivative f'(x) which will give the slope of any point x on the line f(x). Hence, m=f'(-1) since that is the point we are interested in.

f'(x)=(3-5x^2)'=-10x

m=f'(-1)=-10(-1)=10

Now plug m into eq. (1). Solve for b such that x and y go through the point of interest.

y(x)=mx +b

-2=10(-1)+b

8=b

You now have your line that is tangent line at the point (-1,-2).

y(x)=10x +8

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