Question

for the curve x + y^2 - 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. which of the following represents all lines tangent to the curve x + y^2 - y = 1 at x = 1? a. y = -x + 1 and y = -x + 2

Explanation:

Step1: Differentiate the curve implicitly

Differentiate $x + y^{2}-y = 1$ with respect to $x$.
Using the sum - rule and chain - rule, we have $1 + 2y\frac{dy}{dx}-\frac{dy}{dx}=0$.
Rearrange to solve for $\frac{dy}{dx}$: $\frac{dy}{dx}(2y - 1)=-1$, so $\frac{dy}{dx}=\frac{-1}{2y - 1}$.

Step2: Find the y - values when x = 1

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

Step3: Find the slopes of the tangent lines

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

Step4: Use the point - slope form to find the equations of the tangent lines

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