Question

find a function whose second derivative is y''=12x-2 at each point (x,y) on its graph and y=-x+5 is tangent to the graph at the point corresponding to x=1

Answer 1

Start by integrating y'' twice to get function y(x).

$y' = \int (12x - 2) dx = 6x^2 -2x + c \\ \\ y = \int (6x^2 -2x+c )dx= 2x^3-x^2 +cx+d$

Next find value of coefficients 'c' and 'd' using the given tangent line
The slope of tangent line is equal to y' when x=1.

$6x^2 -2x+c = -1, x = 1 \\ \\ 6-2+c = -1 \\ \\ c = -5$

The tangent line intersects the curve y(x) when x = 1.
If x = 1, then y = 4  (From y =-x+5)

$y = 2x^3 -x^2-5x +d , x = 1, y = 4 \\ \\ 2-1-5+d = 4 \\ \\ d = 8$

Final Answer:
$y = 2x^3 -x^2 -5x+8$