Calculus confuses me to no end.... help please?

Question

find the tangent line to the curve y= 2x^2 + 4x - 3 at the point (1,3)

I know that the derivative is 4x + 4, but I don't know where to go from there

Answer 1

The key is to remember that the derivative is the slope of a curve at point x. The tangent line of a curve at a given point is simply a line that passes through that point and has the same slope as the curve does at that point.

You've managed to figure out that the slope is equal to 4x+4, and it's given that x = 1, since the point you're focused on is (1,3)

Just plug in 1 for x to find the slope at that point and you get that it equals 4(1) + 4 or 8.

Now you know the slope of the line and a point that it intersects, so just use the point-slope formula...

y - 3 = 8(x - 1)
y = 8x - 8 + 3
y = 8x -5

And there is your tangent line.

Answer 2

this is not calculus, this is only pre-calc. You take the derivative at that point. So since now you have 4x+4, plug in 1 and you get the answer to be 8. this is the slope of the tangent. You now have y=mx+b where m is 8, now simply plug in 1 and 3 for x and y and solve for b. Simple as that.

Update to other respondents: Dont just give out the answer, teach them how to do it. If they can't do something this simple they will die when they start doing real calc. Get them ready for the future, don't spoon feed them.

Answer 3

You equate 4x=4=0 to find the maxima or minima.

Then you solve for x to get the maxima or minima by equating 4x+4=0.
x= -1.

Try to graph the function, y=2x^2+4x-3 by assigning values of x and solve for y.

You will find whether x = -1 is maxima or minima.

You need graphs to understand more.

Answer 4

The slope is m=4*1+4=8
so
y-3=8(x-1) or y=8x-5