Find an equation for the line tangent to the graph of the fu

[K_Swiss]

Consider the graph of the function f(x) = 5x^2 - 8x.

d) Find an equation for the line tangent to the graph of the function at x = 1.

I found f'(x) = 10x - 8 and at x = 1, the slope is 2

So. . .here's what I have. . .

y = 2x + b
y = 2(1) + b
y = 2 + b

Now what?


Textbook Answer: 2x - y - 5 = 0

 

[skeeter]

you need the y-value of the point of tangency ... what is f(1)?

y = 2 + b
f(1) = 2 + b
b = ?

 

[rogerstein]

Since we're trying to get the constant (the y-intercept) of the straight line, all we need is any value for x on the line and its corresponding y value. Well, k swiss, you'll be delighted to know that we're not in the sometimes uncomfortable postition of having a multitude of choices and not being able to decide which one to pick. No, we know one and only one point on the line and that is the tangent point. We're given that, remember, x=1 in this case. And because Fate is occasionally kind, we have a handy means of determining its y coordinate--the original function, since this is the Peter Sellers of points--multiple roles. So we put on its mustache and insert it in the original function, y=5x^2 -8x to find out its y coordinate-- and discover it's -3. (Y=5-8) So we now temporarily plug our known values for x and y back into our equation under construction, and we have -3=2*1+b. b=-5. Thus y=2*x-5, or as your book has it, 2*x-y-5=0.