Find the equation of the tangent line to y= 5/2x-3 at x=-2

How would i go about solving this problem?

Im not sure how much you have been taught or not, but....

The output of the first derivative function is the slope of the original function at any x point.

So, take the first derivative of the function, input the x value, and out comes the slope. That is m= in the slope point formula.

So now you need a point. If you input the x value into the original function, the output is the y coordinate. Now you have a point. With the slope and a point, you can find the equation of the line.

f(x) = 5/(2x-3)
f ' (x) = -10/((2x-3)^2)
m = f ' (-2) = -10/((2(-2)-3)^2) = -10/(-7)^2 = -10/49
The slope of the line is m = -10/49

So now get the output of the original function
f(-2) = 5/(2(-2)-3) = 5/(-7) = -5/7
So the point is (-2, -5/7)

Now you need an equation of a line. There are many variations, the most common you might see is the slope-intercept formula of y=mx+b, or the standard form of ax+by=c. But I prefer the point-slope formula of (y-y1)=m(x-x1). The reason is you can just input the slope and the point into one equation and get you answer. That is, of course, my personally preference. The others work just as well. Anyway,

We know that m = -10/49 and we know that x1 = -2 and that y1 = -5/7

y - (-5/7) = -10/49(x - (-2))

y + 5/7 = -(10/49)x - 20/49

y = -(10/49)x - 55/49

And there is the equation of the tangent line at x = -2 on f(x)

I hope that made sense. The conceptual aspect of differential calculus is the hard part, not the arithmetic work :)

Good Luck