Please help fast sirs hehe thanks alot

h'(t)=4t³-18t²+3. This equation will provide the slope of the line. So h'(-3)=-4×27-18×9+3=-108-162+3=-267. The slope (coefficient of t) is therefore -267.

The line will have the general form L(t)=-267t+b where we find b by plugging in the coords of the tangent point. When t=-3, h(-3)=81-6×(-27)-9-7=81+162-16=227. So the coords are (-3,227). The line passes through this point.

Therefore 227=-267×(-3)+b, 227=801+b, b=227-801=-574. So L(t)=-267t-574.

Another way to work out the equation is to plug in the point differently (slope-intercept format).

This format is y-y₀=m(x-x₀) where m is the slope, x the independent variable and y the dependent variable. (x₀,y₀) is the tangent point. We have m=-267, and we are using t instead of x and L(t) instead of y.

So we write:

L(t)-227=-267(t-(-3)), L(t)=227-267(t+3)=-267t-574.

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