Tangent to y=sqrt(x)?

Question

Does any tangent to the curve y = sqrt(x) cross the x-axis at x = -1 ? If so, find the equation for the line and the point of tangency. If not, why not?

Answer 1

given y = sqrt(x) = x^(1/2)

y ' = slope of tangent line = 1/(2sqrt(x)) = (1/2)x^(-1/2)

if it crosses the x-axis at x = -1, then the point (-1, 0) will be on a tangent line

the point (x,sqrt(x)) will also be on the tangent line

the slope through these two points must equal the slope from the derivative at that point

(sqrt(x) - 0) / (x - -1) = (1/2) x^(-1/2)

(sqrt(x)) / (x + 1) = 1 / (2sqrt(x))

cross-multiplying:

2x = x + 1

x = 1

so yes, the tangent at the point (1,1) will also pass through the point (-1,0)

slope of this line = 1/2

y = (1/2) x + b

1 = (1/2) 1 + b

b = 1/2

so the equation is y = (1/2) x + 1/2

or x - 2y + 1 = 0

double check: y ' = (1/2) / sqrt(x)

y ' (1) = (1/2) / 1 = 1/2

slope between (1,1) and (-1,0) = 1/2

yep, they match