Final answer:
To find the slope of the tangent line to the parabola y=2x^2-4x+2 at x=-2, we differentiate the equation to get y'=4x-4. Substituting x=-2 gives a slope of -12. The equation of the tangent line is y=-12x-6.
Step-by-step explanation:
The student is asking to find the slope of the tangent line to the parabola y=2x^2-4x+2 at the point where x=-2. To find the slope, we need to take the derivative of the given equation, which will give us the slope of the curve at any point x. The derivative of y=2x^2-4x+2 is y'=4x-4. Substituting x=-2 into this derivative gives us y'=(4*-2)-4=-8-4, so the slope m is -12.
Thus, the equation of the tangent line at the point where x=-2 is y=mx+b. We already have the value of m, which is -12. To find b, we substitute the coordinates of the point into the equation y=mx+b. Since the point lies on the curve, its y-value can be obtained by substituting x=-2 into the original equation, yielding y=2*(-2)^2-4*(-2)+2=8+8+2=18. So, we have 18=-12*(-2)+b which simplifies to 18=24+b. Solving for b gives us b=18-24=-6.
Therefore, the equation of the tangent line at x=-2 is y=-12x-6, and the slope of this tangent line is -12.