Find the equation of tangent line to the curve
Y=2/x^2+x
At the point where x=1
first find y for x=1, you now have a point
take the derivative, sub in x = 1 to get the slope
Now you have a point and a slope, use your method you learned to find the equation of the straight line.
To find the equation of the tangent line to the curve at the point where x=1, we need to find the slope of the tangent line and the coordinates of the point.
Step 1: Find the slope of the tangent line
The slope of the tangent line is equal to the derivative of the curve at the given point. So, we need to find the derivative of the function Y=2/x^2+x.
Differentiating the function Y=2/x^2+x:
Y' = d/dx ( 2/x^2 + x )
= -2*2/x^3 + 1
= -4/x^3 + 1
Step 2: Find the coordinates of the point
Since we are given that x=1, we can substitute this value into the original function to find the corresponding y-coordinate:
Y = 2/1^2 + 1
= 2 + 1
= 3
Therefore, the coordinates of the point where x=1 are (1,3).
Step 3: Use the slope and point to write the equation of the tangent line
We have the slope of the tangent line (-4/x^3 + 1) and the point (1,3).
Using the point-slope form of a linear equation: y - y1 = m(x - x1)
where m is the slope of the line and (x1, y1) are the coordinates of the point, we can substitute the values to get the equation of the tangent line:
y - 3 = (-4/1^3 + 1)(x - 1)
y - 3 = (-4/1 + 1)(x - 1)
y - 3 = (-4 + 1)(x - 1)
y - 3 = -3(x - 1)
Simplifying further:
y - 3 = -3x + 3
To write the equation in the form y = mx + c (slope-intercept form), we isolate y:
y = -3x + 3 + 3
y = -3x + 6
Therefore, the equation of the tangent line to the curve Y=2/x^2+x at the point (1,3) is y = -3x + 6.