Question

for the curve x + y² - y = 1 complete the following parts. a. find equations of all lines tangent to the curve at x = 1. b. graph the tangent lines on the given graph.

Explanation:

Step1: Differentiate the curve implicitly

Differentiate $x + y^{2}-y = 1$ with respect to $x$. The derivative of $x$ is $1$, the derivative of $y^{2}$ using the chain - rule is $2y\frac{dy}{dx}$, and the derivative of $y$ is $\frac{dy}{dx}$, and the derivative of the constant $1$ is $0$. So, $1 + 2y\frac{dy}{dx}-\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the equation $1 + 2y\frac{dy}{dx}-\frac{dy}{dx}=0$ to get $\frac{dy}{dx}(2y - 1)=-1$. Then $\frac{dy}{dx}=\frac{-1}{2y - 1}$.

Step3: Find the $y$ - values when $x = 1$

Substitute $x = 1$ into the original equation $x + y^{2}-y = 1$, we have $1 + y^{2}-y = 1$, which simplifies to $y^{2}-y=0$. Factor out $y$ to get $y(y - 1)=0$. So, $y = 0$ or $y = 1$.

Step4: Find the slopes of the tangent lines

When $y = 0$, $\frac{dy}{dx}=\frac{-1}{2\times0 - 1}=1$. When $y = 1$, $\frac{dy}{dx}=\frac{-1}{2\times1 - 1}=-1$.

Step5: Find the equations of the tangent lines

Using the point - slope form $y - y_{1}=m(x - x_{1})$, when $x_{1}=1,y_{1}=0,m = 1$, the equation is $y-0 = 1\times(x - 1)$, i.e., $y=x - 1$. When $x_{1}=1,y_{1}=1,m=-1$, the equation is $y - 1=-1\times(x - 1)$, i.e., $y=-x+2$.

Answer:

The equations of the tangent lines are $y=x - 1$ and $y=-x + 2$.

For part b, to graph the tangent lines:

1. The line $y=x - 1$ has a $y$ - intercept of $-1$ and a slope of $1$. Plot the point $(0,-1)$ and use the slope to find other points.
2. The line $y=-x + 2$ has a $y$ - intercept of $2$ and a slope of $-1$. Plot the point $(0,2)$ and use the slope to find other points. Then draw these two lines on the same graph as the curve $x + y^{2}-y = 1$.