Find the slope of the function's graph at the given point, then find an equation for the line tangent to the graph there. h(t)=t^3, (-2,-8)
m= then y=
please and thank you, an explanation would be greatly appreciated
m= then y=
please and thank you, an explanation would be greatly appreciated
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h'(t) = 3t^2
At (-2, -6), t = -2
h'(-2) = 3*(-2)^2 = 12
So, m = 12
y = 12(t+2)-8 = 12t + 16
At (-2, -6), t = -2
h'(-2) = 3*(-2)^2 = 12
So, m = 12
y = 12(t+2)-8 = 12t + 16
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h'(t) = 3t²
h'(−2) = 3(−2)² = 12
So tangent line has slope 12 and passes through point (−2, −8)
y + 8 = 12 (t + 2)
y = 12t + 24 − 8
y = 12t + 16
h'(−2) = 3(−2)² = 12
So tangent line has slope 12 and passes through point (−2, −8)
y + 8 = 12 (t + 2)
y = 12t + 24 − 8
y = 12t + 16