Lines tangent to the curve y-1=2x^2+3x pass through the point A(2,-1). Find the x-coordinate(s) of the point(s) of tangency.
The answer is 2+2^1/2, 2-2^1/2, I just don't know how to get there. Please help! Thank you!
To find the x-coordinate(s) of the point(s) of tangency, we need to find the derivative of the curve and solve it for the x-value where the tangent line passes through point A(2,-1).
Step 1: Rewrite the equation of the curve in terms of y:
y - 1 = 2x^2 + 3x
Rewriting it, we get: y = 2x^2 + 3x + 1
Step 2: Find the derivative of the curve:
The derivative of y with respect to x will give us the slope of the tangent line at any given point on the curve.
dy/dx = d/dx(2x^2 + 3x + 1)
= 4x + 3
Step 3: Use the point-slope form of the equation for a line to find the equation of the tangent line passing through point A(2,-1):
y - (-1) = (4x + 3)(x - 2)
Simplify the equation:
y + 1 = 4x^2 - 5x - 6
y = 4x^2 - 5x - 7
Step 4: To find the x-coordinate(s) of the point(s) of tangency, set the tangent line equal to y:
2x^2 + 3x + 1 = 4x^2 - 5x - 7
Rearrange the equation and set it equal to zero:
2x^2 + 8x + 8 = 0
Step 5: Solve the quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = 8, and c = 8.
Substituting these values, we get:
x = (-8 ± √(8^2 - 4 · 2 · 8)) / (2 · 2)
x = (-8 ± √(64 - 64)) / 4
x = (-8 ± √0) / 4
x = (-8 ± 0) / 4
x = -2
Therefore, the x-coordinate(s) of the point(s) of tangency are x = -2.
Please note that in the final answer you mentioned (2 + 2^(1/2), 2 - 2^(1/2)), the square root term seems to be incorrect. The correct solution for the x-coordinate(s) of tangency is x = -2.