find all the points (x,y) of the graph f9x)=2x^2 such that the tangent at (x,y) goes through point (2,0)

Answer 1

One tangent is obviously the line y=0 and the origin is the point needed.

For the other the slope at point x is 4x and that must be equal to y/(x-2)...but y=2x²

Therefore 4x=2x²/(x-2)

You can solve this for x to find the abscissa of the point and then y=2x² gives you the ordinate.

Answer 2

Given the equation f(x) = 2x^2, let's first think about how to find the equation of the tangent line.

To find the slope of the tangent line, we have to take the derivative of the function. Since f'(x) = 4x, that means that the slope of the tangent line at ANY point (x, y) on the graph of f(x) will have the slope of 4x.

So the general form of ANY tangent line of this function (using the point-slope form of a line) will be:

y - y₁ = 4x (x - x₁)

Let's do one more thing before we use the other information we have. Since we know that for the given function, every "y" value can be calculated as 2x^2, we can substitute that in the equation.

So that gives us: 2x^2 - y₁ = 4x (x - x₁). (you'll see why this is helpful in the next step)

Now, what we want is for this line to pass through (2, 0). Therefore, we substitute that in the (x₁, y₁) values of our line:

2x^2 - 0 = 4x (x - 2). [See how now we have an equation that is all in one variable, so we can solve for x]

And now, we solve for x:

2x^2 = 4x^2 - 8x

0 = 2x^2 - 8x

0 = 2x(x - 4)

x = 0, 4

We substitute these x values back into our original function to find the coordinates (x, y):

x = 0 --> f(0) = 2(0)^2 = 0 --> (0, 0)

x = 4 --> f(4) = 2(4)^2 = 32 --> (4, 32)

So your two points are (0, 0) and (4, 32)