Derivative equation tangent line to f(x) = 3x(5x^2 + 1) parallel to y = 8x + 9

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Alexmcom

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Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1)
that are parallel to the line y=8x+9



Okay so never done this before but I'm stuck and hoping someone could guide me. I find the derivative of 3x(5x^2+1) which is 45x^3+3

I get the y formula and find the derivative which is 8

I want to find the slope so i plug in 8 on the left side of 45x^3+3 which should look like this. 8=45x^3+3


Bring 3 to the other side making it 5

So its 5=45x^3

divide each side so its 1=9x^3 which I find very odd since I believe here I've already messed up.

Then I plug in 1=9x^3 to the original derivative and here I get lost. 3(9x^3)(5(9x^3)^2+1)


Can someone guide me???
 
Alexmcom said:
Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1)
that are parallel to the line y=8x+9



Okay so never done this before but I'm stuck and hoping someone could guide me. I find the derivative of 3x(5x^2+1) which is 45x^3+3

I get the y formula and find the derivative which is 8

I want to find the slope so i plug in 8 on the left side of 45x^3+3 which should look like this. 8=45x^3+3


Bring 3 to the other side making it 5

So its 5=45x^3

divide each side so its 1=9x^3 which I find very odd since I believe here I've already messed up.

Then I plug in 1=9x^3 to the original derivative and here I get lost. 3(9x^3)(5(9x^3)^2+1)


Can someone guide me???
Click to expand...

How did you get that?
 
By finding the derivative of f(x) and the y axis.


But I know I messed up hard and hoping I could get some guidance
 
After a little research apparently I am suppose to use the quadratic formula for this one, is this correct?
 
Alexmcom said:
After a little research apparently I am suppose to use the quadratic formula for this one, is this correct?
Click to expand...
I do not see any need of using quadratic here. You need to calculate the tangent point on the curve(where the tangent line touches the curve).
 
Alexmcom said:
Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1)
that are parallel to the line y=8x+9
Okay so never done this before but I'm stuck and hoping someone could guide me. I find the derivative of 3x(5x^2+1) which is 45x^3+3 Incorrect! So its 5=45x^3 This is useless and incorrect.
Click to expand...

This is correct:
\(\displaystyle \begin{align*}45x^2+3&=8 \\45x^2&= 5\\x^2&=9^{-1}\\x&=\pm 3^{-1} \end{align*}\)
 
Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1)
that are parallel to the line y=8x+9



Okay so never done this before but I'm stuck and hoping someone could guide me. I find the derivative of 3x(5x^2+1) which is 45x^3+3


Can someone guide me???
Subhotosh Khan said:
How did you get that?
Click to expand...
Actually the 45 is correct, it is the power 3 that confuses me. But then again I am always confused.
 
Alexmcom said:
Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1)
that are parallel to the line y=8x+9



Okay so never done this before but I'm stuck and hoping someone could guide me. I find the derivative of 3x(5x^2+1) which is 45x^3+3

I get the y formula and find the derivative which is 8

I want to find the slope so i plug in 8 on the left side of 45x^3+3 which should look like this. 8=45x^3+3


Bring 3 to the other side making it 5

So its 5=45x^3

divide each side so its 1=9x^3 which I find very odd since I believe here I've already messed up.

Then I plug in 1=9x^3 to the original derivative and here I get lost. 3(9x^3)(5(9x^3)^2+1)


Can someone guide me???
Click to expand...


If tangents are paralel to \(\displaystyle y=8x+9 \) , then slope of tangents must be equal to the slope of that line. So, tangents should be: \(\displaystyle y=8x + a \). Since first derivative represents the slope of tangents on a graph, you must find the first derivative of \(\displaystyle f(x)=3x(5x^2+1) \). When you have \(\displaystyle f'(x) \), you can now find x-axis coordinates of touching points of required tangents as follows: \(\displaystyle f'(x)=8 \) . To get a in the equation of tangents you must plug \(\displaystyle y=8x+a \) into the equation of your graph: \(\displaystyle 8x+a=3x(5x^2+1) \) . Now you have \(\displaystyle 15x^3 -5x -a=0 \) .
Now plug x values that you got from \(\displaystyle f'(x)=8 \) in this last equation and you will get a. Your tangents are \(\displaystyle y_1=8x+a_1 \, y_2=8x+ a_2 \) .

P.S. You got the derivative wrong. It's not \(\displaystyle 45x^3+3 \). It's: \(\displaystyle 45x^2 +3 \) .
 
Last edited:
Johulus said:
If tangents are paralel to \(\displaystyle y=8x+9 \) , then slope of tangents must be equal to the slope of that line. So, tangents should be: \(\displaystyle y=8x + a \). Since first derivative represents the slope of tangents on a graph, you must find the first derivative of \(\displaystyle f(x)=3x(5x^2+1) \). When you have \(\displaystyle f'(x) \), you can now find x-axis coordinates of touching points of required tangents as follows: \(\displaystyle f'(x)=8 \) . To get a in the equation of tangents you must plug \(\displaystyle y=8x+a \) into the equation of your graph: \(\displaystyle 8x+a=3x(5x^2+1) \) . Now you have \(\displaystyle 15x^3 -5x -a=0 \) .
Now plug x values that you got from \(\displaystyle f'(x)=8 \) in this last equation and you will get a. Your tangents are \(\displaystyle y_1=8x+a_1 \, y_2=8x+ a_2 \) .

P.S. You got the derivative wrong. It's not \(\displaystyle 45x^3+3 \). It's: \(\displaystyle 45x^2 +3 \) .
Click to expand...




I plugged in 8 into the axis and got 7680-40-a=0

Subtracted 40 from 7680 which equals 7640-a

I really find this an odd number and clearly as again I have messed up.
 
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