Question

Find the point on the curve Y=x^2-2x+3 where the tangent is parallel to y-axis​

Answer 1

Answer:

Step-by-step explanation:

The tangent cannot be parallel to the y-axis.

since this is a parabola.

So, tangent can only be parallel to the x-axis.

For tht point, u have to find the minima => dy/dx = 0

so dy/dx = 2x-2

so x=1

then when x=1, y = 1^2 - 2*1 +3 = 2

so the required point is (1,2)

Answer 2

Answer:

Let's rewrite y=x2–2x+3 in a useful form.

First, in order to line up with the intended parallel line, we reserve x ; now our equation looks like:

y=(x2–3x+3)+x .

Now we compete the square on what's left in the parentheses:

y=(x^2–3x+9/4)+x+3/4

y=(x–3/2)^2+x+3/4

Since the part in parentheses describes a parabola with vertex at x=3/2 , the rest of it describes the line tangent to y=x^2–2x+3 at x=3/2 .

That is, the line y=x+3/4 is tangent to y=x^2–2x+3 when x=3/2 . And, of course, we note that when x=−3/4 , x+3/4=0 , so the tangent line crosses the x -axis at (−3/4,0) .