Final answer:
To find the equation of tangents to the curve y=2x²+3, differentiate the curve to get the slope, then use the points (2,3) and (2,-7) with the slope to write the equation of the tangents, resulting in y=8x-13 and y=8x-23, respectively.
Step-by-step explanation:
To find the equations of the tangents to the curve y=2x²+3 that pass through a given point, we must first determine the slope of the tangent lines to the curve at any point by differentiating the equation of the curve. The derivative is y'=4x, which gives us the slope at any point (x, y) on the curve.
Finding the Tangent Through Point (2,3)
The slope of the tangent at x=2 is y'=4(2)=8. Since the tangent passes through (2,3), the equation of the tangent is y-3=8(x-2). Simplifying this gives y=8x-13.
Finding the Tangent Through Point (2,-7)
The slope of the tangent at x=2 is the same, so y' = 8. Using the point (2,-7), the equation is y+7=8(x-2). Simplifying, we have y=8x-23.