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Q32.

Expert-verifiedFound in: Page 424

Book edition
Middle English Edition

Author(s)
Carter

Pages
804 pages

ISBN
9780079039903

**Use the equation $x=3{y}^{2}+4y+1$ to find the y-intercept.**

The *y-*intercept of the equation $x=3{y}^{2}+4y+1$ is** $\left(0,-\frac{1}{3}\right)\text{}\mathbf{and}\text{}\left(0,-1\right)$.**

The given equation is $x=3{y}^{2}+4y+1$.

The *y-*intercept of the equation $x=3{y}^{2}+4y+1$ can be obtained by plugging $x=0$ in $x=3{y}^{2}+4y+1$. Therefore,

$\begin{array}{c}x=3{y}^{2}+4y+1\left(\text{Given}\right)\\ \left(0\right)=3{y}^{2}+4y+\mathrm{1....}\left(\text{Plugging}x=0\right)\\ 3{y}^{2}+4y+1=0\\ \left(3y+1\right)\left(y+1\right)=0\\ 3y+1=0\text{or}y+1=0\\ y=-\frac{1}{3}\text{or}y=-1\end{array}$

The *y-*intercept of the equation $x=3{y}^{2}+4y+1$ is $\left(0,-\frac{1}{3}\right)\text{and}\left(0,-1\right)$.

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