For the curve y = 3x^4 - 6x^2 + 2, find the slope at the point whose x-coordinate is 2 and an equation for the tangent line there.

Step-by-step solution

1. 1

   Differentiate the curve. Term by term, $\frac{dy}{dx}=12x^3-12x$. The derivative is the slope function: feeding it an $x$-value returns the slope of the tangent line there.

2. 2

   Evaluate the slope at $x=2$. $12(2)^3-12(2)=12(8)-24=96-24=72$, so the tangent has slope $m=72$.

3. 3

   Find the actual point of tangency. The worksheet labels the point $(2,-2)$, but substituting gives $y(2)=3(16)-6(4)+2=48-24+2=26$. The curve passes through $(2,26)$; the printed $y$-value is a misprint, and using it would place the line in the wrong position.

4. 4

   Apply point-slope form. With $m=72$ and $(x_1,y_1)=(2,26)$: $y-26=72(x-2)$.

5. 5

   Expand to slope-intercept form. $y-26=72x-144$, hence $y=72x-118$.

6. 6

   Verify. At $x=2$ the line gives $144-118=26$, which matches the curve, and its slope $72$ matches the derivative — both conditions a tangent line must satisfy.