Question

How do you write $Y=-6{{\left( x-2 \right)}^{2}}-9$ in standard form?

Answer 1

Hint: As we know that the standard form of a quadratic equation is given as $a{{x}^{2}}+bx+c=0$ where, x is an unknown variable and a, b, c are real numbers. So to write the given equation in standard form we just simplify the given equation. We will solve the operators given in the equation to get the desired answer.

Complete step by step solution:
We have been given an equation $Y=-6{{\left( x-2 \right)}^{2}}-9$.
We have to write the given equation in standard form.
Now, we know that the standard form of an equation is $a{{x}^{2}}+bx+c=0$, where x is an unknown variable and a, b, c are real numbers. We have to convert the given equation in standard form by simplifying the equation.
Let us simplify the operators given in the equation to write it in standard form. Then we will get
$\begin{align}
  & \Rightarrow Y=-6{{\left( x-2 \right)}^{2}}-9 \\
 & \Rightarrow Y=-6\left( {{x}^{2}}+{{2}^{2}}-2\times 2x \right)-9 \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow Y=-6\left( {{x}^{2}}+4-4x \right)-9 \\
 & \Rightarrow Y=-6{{x}^{2}}-24+24x-9 \\
 & \Rightarrow Y=-6{{x}^{2}}+24x-33 \\
\end{align}$
Hence above is the required standard form of the given equation.

Note: We can further simplify the obtained equation by dividing the whole equation by 3 then we will get
$\Rightarrow Y=-\dfrac{6{{x}^{2}}}{3}+\dfrac{24x}{3}-\dfrac{33}{3}$
Now, simplifying the above obtained equation we will get
$\Rightarrow Y=-2{{x}^{2}}+8x-11$
If $a=0$ in the equation $a{{x}^{2}}+bx+c=0$ then the equation is a linear equation not a quadratic equation. We have methods like factoring, completing the square method and quadratic formula method to solve the quadratic equation.