Help needed.

[SurnameLong]

Find the slope of the curve and the equation of the tangent to the curve at each of the points indicated.
(a) y=1/4x³ at x=2
I differentiated y to get 3/4x². (3/4)(2²)=3. I subbed this in to y-3=3/4(x-2). 4y=3x-15. I know this I wrong yet I don't know how to proceed.
The answer is 3x-y=4.

 

[ksdhart]

Okay, so it appears you're solving the problem (1/4)x^3 or 0.25x^3, correct? If that's the case, the derivative is (3/4)x^2. Evaluated at x=2, the slope at that point is 3. Where you went wrong is the equation for a line tangent to a point is this:

y-y₀ = m(x-x₀)

Remember what each of those terms means. y₀ means the y-value at the point in question. So evaluate y=(1/4)x^3 when x=2. m is the slope of the line at the point. You solved for that by taking the derivative. And finally x₀​ is the x-value at the point. You know that because it was given to you. So, plug in all the values and you'll have the equation.

 

[greg1313]

Find the slope of the curve and the equation of the tangent to the curve at each of the points indicated.
(a) y=1/4x³ at x=2
I differentiated y to get 3/4x². (3/4)(2²)=3. <-------- This is the slope.

I subbed this in to y-3=3/4(x-2). 4y=3x-15.

I know this I wrong yet I don't know how to proceed.
The answer is 3x-y=4. Ax + By = C \(\displaystyle \ \ \) form

y = (1/4)x<sup>3

x = 2

y = (1/4)(2)3</sup>

y = (1/4)(8)

y = 2

The point of tangency is (2, 2), and the slope, m, is 3.

\(\displaystyle y \ - \ y_1 \ = \ m(x \ - \ x_1)\)

y - 2 = 3(x - 2)


Continue from there.

 

[Bob Brown MSEE]

No derivative required

Given:
y = (1/4) x³

Translate so origin is at the desired point (2,2)
(y+2) = (1/4) (x+2)³


Expand in standard form
y = (1/4) x³ + (1/2)x² + 3x
Drop all but linear term
y = 3x
Translate back to (2,2)
(y-2) = 3(x-2)
y = 3x - 4

Slope is 3

 

[SurnameLong]

Thanks for the help.

I really appreciate it.