If tangents to the curve defined by y=x^2+4 ln x are parallel to the line defined by by
y-6x+3=0, determine the points where the tangents touch the curve.
y’=2x+4/x
2x+4/x=6
What do I do next?
you got the calculus right -- what bothers you about the algebra?
You have the slope of the parallel line: m=6
So, that means you need
2x + 4/x = 6
2x^2-6x+4 = 0
2(x-1)(x-2) = 0
So the lines are tangent at (1,1) and (2,4+4ln2)
To find the points where the tangents touch the curve, we need to solve the equation 2x + 4/x = 6.
To begin, let's rearrange the equation and get rid of the fraction by multiplying every term by x:
2x^2 + 4 = 6x
Rearrange the equation to set it equal to zero:
2x^2 - 6x + 4 = 0
Now, using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = -6, and c = 4. Plugging these values into the formula, we get:
x = (-(-6) ± √((-6)^2 - 4 * 2 * 4)) / 2 * 2
x = (6 ± √(36 - 32)) / 4
x = (6 ± √4) / 4
Now we can simplify the expression further:
x = (6 ± 2) / 4
This gives us two possible values for x:
x1 = (6 + 2) / 4 = 8 / 4 = 2
x2 = (6 - 2) / 4 = 4 / 4 = 1
So, we have found two x-values, x = 1 and x = 2. To find the corresponding y-values on the curve, we substitute these values back into the original equation y = x^2 + 4ln(x):
For x = 1:
y = 1^2 + 4ln(1) = 1 + 4(0) = 1
Therefore, one point where the tangent touches the curve is (1, 1).
For x = 2:
y = 2^2 + 4ln(2) = 4 + 4ln(2)
So the other point where the tangent touches the curve is (2, 4 + 4ln(2)).
Therefore, the points where the tangents touch the curve are (1, 1) and (2, 4 + 4ln(2)).