Find the equation of a tangent line of a graph pf y=x^2+4lnx, which is parallel to y=6x-3?
I know parallel lines have the same slope so the slope is 6 but I do not know how to find the b value.
Well you need to find where on that nasty function the slope is 6. Then at that (x,y) you write the y = 6 x + b that goes through that point.
how do I do this?
y = x^2 + 4 ln x
dy/dx = slope = 2 x +4/x
when is that slope 6?
6 = 2 x + 4/x
I get at x = 1 and at x = 2
well 1 is easy because ln 1 = 0. It just said "a" tangent line, not all tangent lines.
then y = 1
so I am going to use (1,1)
actually it's not so bad
the derivative of your function is
2x + 4/x
setting this equal to 6 gave me a quadratic which factored nicely and had roots of
x = 1 or x=2
if x=1 then y = 1+4ln1
y=1 ,
so one case is m=6, point (1,1)
case 2
x=2, then y = 4 + 4ln2
so m=6, point (2,4+4ln2)
etc
All set now?
I understand what you are saying but how do you do the algebra foe 6=2x+4/x I seem to have forgotten my algebra.
No i need help understanding the algebra behind solving 2x+4/x=6
6=2x+4/x
multiply both sides by x
6 x = 2 x^2 + 4
2 x^2 - 6 x + 4 = 0
1 x^2 - 3 x + 2 = 0
factor
(x-1)(x-2) = 0
x = 1
x = 2 are solutions.
I used 1 so the point I could take was (1,1)
To find the equation of the tangent line that is parallel to y=6x-3, we need to determine the point on the graph of y=x^2+4lnx where the tangent line will intersect.
To do this, we can use the fact that the slope of the tangent line is equal to the derivative of the function at that point. Let's find the derivative of y=x^2+4lnx using the rules of differentiation.
First, differentiate each term separately:
dy/dx (x^2) = 2x
dy/dx (4lnx) = 4 (1/x) = 4/x
Then, combine the results:
dy/dx = 2x + 4/x
Now, we can set the derivative equal to the given slope, which is 6:
2x + 4/x = 6
To solve this equation for x, we can rearrange it to a quadratic equation:
2x^2 - 6x + 4 = 0
Factorizing the quadratic equation, we get:
2(x - 1)(x - 2) = 0
Setting each factor equal to zero, we find two possible values for x: x = 1 and x = 2.
Now, substitute these values back into the original equation to find the corresponding y-values:
When x = 1, y = (1)^2 + 4ln(1) = 1 + 4(0) = 1
When x = 2, y = (2)^2 + 4ln(2) = 4 + 4ln2
So, we have two points on the graph: (1, 1) and (2, 4 + 4ln2).
Given that the tangent line is parallel to y=6x-3 (which has a slope of 6), we can use the point-slope form of a linear equation to write the equation of the tangent line.
Using the point (1, 1), we have:
y - y1 = m(x - x1),
where m is the slope and (x1, y1) is the given point.
Replacing m with 6 and substituting the values of x1 and y1, we get:
y - 1 = 6(x - 1)
Simplifying the equation will give us the equation of the tangent line in the form y = mx + b:
y - 1 = 6x - 6
y = 6x - 5
Therefore, the equation of the tangent line that is parallel to y=6x-3 and intersects the graph of y=x^2+4lnx at the point (1, 1) is y = 6x - 5.