Find the y-intercept of the tangent line to the curve $y = \sqrt { x + 3 }$ at the point (1, 2).
A) $\frac { 3 } { 4 }$
B) $\frac { 7 } { 4 }$
C) $\frac { 1 } { 4 }$
D)2
E) $\frac { 1 } { 2 }$
F)1
G) $\frac { 3 } { 2 }$
H) $\frac { 5 } { 4 }$

Question

Find the y-intercept of the tangent line to the curve $y = \sqrt { x + 3 }$ at the point (1, 2).
A) $\frac { 3 } { 4 }$ 
B) $\frac { 7 } { 4 }$
C) $\frac { 1 } { 4 }$
D)2
E) $\frac { 1 } { 2 }$ 
F)1
G) $\frac { 3 } { 2 }$ 
H) $\frac { 5 } { 4 }$

Quizplus · Accepted Answer

To find the y-intercept of the tangent line to the curve at a given point, first, find the derivative of the curve to get the slope of the tangent line at that point. The derivative of $y = \sqrt{x + 3}$ is $y' = \frac{1}{2\sqrt{x + 3}}$. At $x = 1$, $y' = \frac{1}{2\sqrt{1 + 3}} = \frac{1}{4}$. The equation of the tangent line in point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point on the curve. Substituting $m = \frac{1}{4}$, $x_1 = 1$, and $y_1 = 2$, we get $y - 2 = \frac{1}{4}(x - 1)$. To find the y-intercept, set $x = 0$: $y - 2 = \frac{1}{4}(0 - 1)$, so $y = 2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$.

Find the Y-Intercept of the Tangent Line to the Curve y=x+3y = \sqrt { x + 3 }y=x+3​

Find the Y-Intercept of the Tangent Line to the Curve $y = \sqrt { x + 3 }$