To find the value of h for which y = -x + 6 is a tangent to the curve y = 5x^2 + 6x - 3h, we need to find the point of tangency. The point of tangency is the point at which the two equations have the same x and y values. So, we can set the two equations equal to each other and solve for x and y.
y = -x + 6
y = 5x^2 + 6x - 3h
-x + 6 = 5x^2 + 6x - 3h
Now we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
where a = 5, b = 6, and c = -3h - 6
x = (-6 ± √(6^2 - 4(5)(-3h-6))) / 2(5)
x = (-6 ± √(36 + 60h + 24)) / 10
x = (-6 ± √(60h + 60)) / 10
Now we can substitute this value of x back into one of the original equations to find y:
y = -x + 6
y = -(-6 ± √(60h + 60)) / 10 + 6
y = 6 ± √(60h + 60) / 10
Now we can substitute this value of x and y back into the other equation:
y = 5x^2 + 6x - 3h
6 ± √(60h + 60) / 10 = 5x^2 + 6x - 3h
Now we can use the fact that the slope of the tangent line is equal to the derivative of the function at the point of tangency.
So, the derivative of y = 5x^2 + 6x - 3h is:
dy/dx = 10x + 6
So, we can substitute x and y values back into the derivative and get the slope of the tangent line:
10x + 6 = -1
10x = -7
x = -7/10
Now, substituting this value of x back into one of the original equations to find y:
y = -x + 6
y = -(-7/10) + 6
y = 7/10 + 6
y = 61/10
So, the point of tangency is (x,y) = (-7/10, 61/10)
Finally, we can substitute this point of tangency into one of the original equations and solve for h.
61/10 = 5(-7/10)^2 + 6(-7/10) - 3h
h = (61/10 + 49/100 + 21/100 - 3h) / -3
h = (61/10 + 70/100) / -3
h = -(431/30)
Therefore, h = -(431/30) for y = -x + 6 is a tangent to the curve y = 5x^2 + 6x - 3h