Let f(t)=8t^2+4t+1. Find the value of t for which the tangent line to the graph of f(t) has slope 1.

Question

xishem · Answer

$$f(t)=8t^2+4t+1$$$$f'(t)=16t+4$$$$1=16t+4$$$$t=\frac{-3}{16}$$

anonymous · Answer

god i feel so slow. a lot of these are the same just phrased differently. thanks for all the help.

xishem · Answer

Just remember this, and maybe you'll be able to see how a lot of the problems are very similar in nature, just worded differently:

The derivative a function, f(x), is a function to find the slope of the tangent line at any x value. The slope of the tangent line is the instantaneous rate of change for the original function, f(x).

Whenever the question asks anything about the tangent line of a function, the first step is almost always to differentiate. Just in case, in any opening calculus class, differentiate first, and then work from there.

Just try to remember what the derivative a function actually is. Don't just memorize how to use it.

anonymous · Answer

theres actually another problem i posted before this one. could you look at it too? i got -1 but thats incorrect according to my hw site.