Answer:
y = 4x + 14
Step-by-step explanation:
To find the equation of the tangent to the parabola (x + 6)² = 3(y - 2) at the given point (0, 14), start by determining the slope of the tangent at any point through differentiation of the equation with respect to x.
Differentiate (x + 6)² = 3(y - 2) with respect to x:
\(\dfrac{d}{dx}((x + 6)^2) = \dfrac{d}{dx}(3(y - 2))\\\\\\\dfrac{d}{dx}(x^2+12x+36) = \dfrac{d}{dx}(3y-6)\\\\\\2x+12=3\dfrac{dy}{dx}\\\\\\\dfrac{dy}{dx}=\dfrac{2}{3}x+4\)
Substitute x = 0 into dy/dx to find the slope of the tangent at point (0, 14):
\(\dfrac{dy}{dx}=\dfrac{2}{3}(0)+4\\\\\\\dfrac{dy}{dx}=4\)
So, the slope of the tangent at point P is m = 4.
Now, substitute the slope (m = 4) and the coordinates of point P (0, 14) into the point-slope equation and simplify:
\(y-y_1=m(x-x_1)\\\\\\y - 14 = 4(x - 0)\\\\\\y-14=4x\\\\\\y=4x+14\)
Therefore, the equation of the tangent to the parabola at point P(0, 14) is:
\(\LARGE\boxed{\boxed{y = 4x + 14}}\)
