Q15. - Algebra 2

Question

Write the equation $x=3y^2+5y-9$ in standard form.

Step-by-Step Solution

Answer

The required form of the equation in standard form $x=a{\left(y-k\right)}^2+h$ is $x=3{\left(y+\frac{5}{6}\right)}^2-\frac{133}{12}$ where $a=3,h=-\frac{133}{12} \text{and }k=-\frac{5}{6}$.

Step 1. Write down the given information.

The given equation is $x=3y^2+5y-9$.

Step 2. Calculation.

The given equation $x=3y^2+5y-9$ is converted to standard form $x=a{\left(y-k\right)}^2+h$ as:

 $\begin{array}{c}x=3y^2+5y-9\\ =3\left(y^2+\frac{5}{3}y-3\right)\\ =3\left(y^2+\frac{5}{3}y-3+{\left(\frac{5}{6}\right)}^2-{\left(\frac{5}{6}\right)}^2\right)\mathrm{....}\left(\begin{array}{l}\text{ Add and }\\ \text{subtract }{\left(\frac{5}{6}\right)}^2\end{array}\right)\\ =3\left(y^2+\frac{5}{3}y+{\left(\frac{5}{6}\right)}^2-\left(3+{\left(\frac{5}{6}\right)}^2\right)\right)\\ =3{\left(y+\frac{5}{6}\right)}^2-\frac{133}{12}\end{array}$

Here $a=3,h=-\frac{133}{12} \text{and }k=-\frac{5}{6}$.

Step 3. Conclusion.

The required form of the given equation is $x=3{\left(y+\frac{5}{6}\right)}^2-\frac{133}{12}$where $a=3,h=-\frac{133}{12} \text{and }k=-\frac{5}{6}$.