To find the equations of the tangent lines to the curve y = x^2 + 1 that are parallel to the line 2y = x + 2, we begin by determining the slope of the given line.
The equation 2y = x + 2 can be rewritten in slope-intercept form (y = mx + b) as follows:
Divide both sides by 2:
y = (1/2)x + 1
From this equation, we can see that the slope (m) of the line is 1/2.
Next, we need to find the points on the curve y = x^2 + 1 where the slopes of the tangent lines are also 1/2. To do this, we start by finding the derivative of the curve:
y’ = d/dx (x^2 + 1) = 2x
We set the derivative equal to the slope of the line:
2x = 1/2
x = 1/4
Now that we have the x-coordinate of the point where the tangent line is parallel to the given line, we can find the corresponding y-coordinate by substituting x = 1/4 into the original curve equation:
y = (1/4)^2 + 1 = 1/16 + 1 = 17/16
Thus, the point of tangency is (1/4, 17/16).
Next, we’ll write the equation of the tangent line using the point-slope form of a line equation:
y – y_1 = m(x – x_1)
y – 17/16 = 1/2(x – 1/4)
Simplifying this gives:
y – 17/16 = 1/2x – 1/8
y = 1/2x – 1/8 + 17/16
y = 1/2x + 7/16
Therefore, one of the equations of the tangent line is y = 1/2x + 7/16.
Now, we should check if there are other points on the curve that yield tangent lines with the same slope. Since the second derivative gives us the curvature of the function, and because it’s a quadratic function (parabola), the tangent line we found is the only one for the given slope.
In summary, the equation of the tangent line to the curve y = x^2 + 1 that is parallel to the line 2y = x + 2 is:
y = 1/2x + 7/16