Calulus. Please explain answer!
y = 2√x - x + 7
y = 2 * x^(1/2) - x + 7
y' = x^(-1/2) - 1
y' = (1/√x) - 1 ←←[gives the slope of the tangent at any point on the curve]
so when x = 4, y' = (1/2) - 1 = -1/2
so the tangent touches the curve at (4, 7) with slope -1/2 and has the form y = mx + b
subsing:
7 = (-1/2) * 4 + b
b = 9
so the required equation is y = (-1/2)x + 9
and that can also be written as x + 2y = 18
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y = 2√x - x + 7
y = 2 * x^(1/2) - x + 7
y' = x^(-1/2) - 1
y' = (1/√x) - 1 ←←[gives the slope of the tangent at any point on the curve]
so when x = 4, y' = (1/2) - 1 = -1/2
so the tangent touches the curve at (4, 7) with slope -1/2 and has the form y = mx + b
subsing:
7 = (-1/2) * 4 + b
b = 9
so the required equation is y = (-1/2)x + 9
and that can also be written as x + 2y = 18