Question

How do you write $y=-3{{x}^{2}}+5x-2$ in vertex form?

Answer 1

Hint: The vertex form of the equation of the form $y=a{{x}^{2}}+bx+c$ is given by $y=a{{\left( x-h \right)}^{2}}+k.$ In this equation, $h$ is the $x-$coordinate of the vertex. Also, $k$ is the $y-$coordinate of the vertex.

Complete step by step solution:
Consider the equation of the form $y=a{{x}^{2}}+bx+c.$
The vertex form of the above equation is given by $y=a{{\left( x-h \right)}^{2}}+k$ where $a$ is the coefficient of ${{x}^{2}},$ $h$ is the $x-$coordinate of the vertex and $k$ is the $y-$coordinate of the vertex.
In this equation, $h$ is obtained by $h=\dfrac{-b}{2a}$ and $k$ is obtained by applying the value of $h$ in the above equation for $k$ is the $y-$coordinate corresponding to the $x-$coordinate $h.$
Let us consider the given equation $y=-3{{x}^{2}}+5x-2.$
We are asked to find the vertex form of the given equation.
So, let us write the coefficients first.
The coefficient of the term ${{x}^{2}}$ is $-3.$ That is, $a=-3.$
The coefficient of the term $x$ is $5.$ That is $b=5.$
Now, we are going to find the value of the $x-$coordinate.
When we apply the values in the formula for $h,$ we will get $h=\dfrac{-b}{2a}=\dfrac{-5}{2\cdot \left( -3 \right)}.$
Now we will get $h=\dfrac{-5}{-6}=\dfrac{5}{6}.$ Therefore, the value of $x-$coordinate of the vertex is $h=\dfrac{5}{6}.$
We are going to apply this value in the given equation to get the $y-$coordinate $k$ of the vertex.
So, $k=-3{{\left( \dfrac{5}{6} \right)}^{2}}+5\left( \dfrac{5}{6} \right)-2.$
Now we will get $k=-3\dfrac{25}{36}+5\dfrac{5}{6}-2.$
That is, $k=-3\dfrac{25}{36}+\dfrac{25}{6}-2.$
Now, cancel the common factor $3$ in the first summand from the numerator and the denominator to get $k=\dfrac{-25}{12}+\dfrac{25}{6}-2.$
Make the denominators the same, $k=\dfrac{-25}{12}+\dfrac{2\times 25}{2\times 6}-\dfrac{2\times 12}{12}=\dfrac{-25}{12}+\dfrac{50}{12}-\dfrac{24}{12}$
Now we will get $k=\dfrac{25}{12}-\dfrac{24}{12}=\dfrac{1}{12}.$
Now the vertex form of the equation is obtained by substituting these values in the equation $y=a{{\left( x-h \right)}^{2}}+k$ as, \[y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12}.\]
Hence the vertex form is \[y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12}.\]

Note: We can simplify the vertex form \[y=-3{{\left( x-\dfrac{5}{6} \right)}^{2}}+\dfrac{1}{12}\] as follows:
Now we get $y=-3\left( {{x}^{2}}-2x\dfrac{5}{6}+{{\left( \dfrac{5}{6} \right)}^{2}} \right)+\dfrac{1}{12},$ since ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}.$
Now we are cancelling the common factors from the numerator and the denominator of the second term and squaring the third term, $y=-3\left( {{x}^{2}}-\dfrac{5}{3}x+\dfrac{25}{36} \right)+\dfrac{1}{12}.$
Take $-3$ inside the bracket as $y=-3{{x}^{2}}+3\dfrac{5}{3}-3\dfrac{25}{36}+\dfrac{1}{12}.$
We will get $y=-3{{x}^{2}}+5x-\dfrac{25}{12}+\dfrac{1}{12}=-3{{x}^{2}}+5x-\dfrac{24}{12}.$