JK has endpoints at -12 and 4. What is the coordinate of its midpoint
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<h3>Answer: y = x+1</h3>
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Explanation:
f(x) = x^3 - 2x + 3
f ' (x) = 3x^2 - 2 ..... apply the power rule
f ' (1) = 3(1)^2 - 2 ... plug in x coordinate of given point
f ' (1) = 1
If x = 1 is plugged into the derivative function, then we get the output 1. This means the slope of the tangent line at (1,2) is m = 1. It's just a coincidence that the x input value is the same as the slope m value.
Now apply point slope form to find the equation of the tangent line
y - y1 = m(x - x1)
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x - 1 + 2
y = x + 1 is the equation of the tangent line.
The graph is shown below. I used GeoGebra to make the graph.