Question

How do you rewrite $y=4{{x}^{2}}-4x-5$ in vertex form? What is the vertex and the axis of symmetry?

Answer 1

In the given question, we have been asked to write a given equation in the vertex form. In order to write the equation in a vertex form, we need to use the vertex form of the equation; i.e.$y=a{{\left( x-h \right)}^{2}}+k$ where, $a$ equals to the coefficient of ${{x}^{2}}$, ‘h’ is the x-coordinate of the vertex, ‘k’ is the y-coordinate of the vertex. First we need to solve for the coordinate of the vertex and then putting all the values we will get our equation in vertex form. The x-coordinate we will find, it will be the line of symmetry.

Formula used:
The vertex form of the equation; i.e.
$y=a{{\left( x-h \right)}^{2}}+k$
Where,
$a$ equals to the coefficient of ${{x}^{2}}$.
‘h’ is the x-coordinate of the vertex.
‘k’ is the y-coordinate of the vertex.

Complete step by step solution:
We have given that,
$\Rightarrow y=4{{x}^{2}}-4x-5$
Using the vertex form of the equation; i.e.
$y=a{{\left( x-h \right)}^{2}}+k$
Where,
$a$ equals to the coefficient of ${{x}^{2}}$.
‘h’ is the x-coordinate of the vertex.
‘k’ is the y-coordinate of the vertex.
Vertex of the given equation $y={{x}^{2}}+14x+29$ is;
Now, solving for the value of ‘x’ and ‘y’,
$\Rightarrow y=4{{x}^{2}}-4x-5$
Here ‘a’ is the coefficient of first term and b is the coefficient of ‘x’.
$\Rightarrow x=\dfrac{-b}{2a}=\dfrac{-\left( -4 \right)}{2\times 4}=\dfrac{4}{8}=\dfrac{1}{2}$
$\Rightarrow x=\dfrac{1}{2}$
Now, solving for the value of ‘y’.
Putting $x=\dfrac{1}{2}$ in the given equation, we get
$\Rightarrow y=4\times {{\left( \dfrac{1}{2} \right)}^{2}}-\left( 4\times \dfrac{1}{2} \right)-5$
Simplifying the above equation, we get
$\Rightarrow y=4\times \left( \dfrac{1}{4} \right)-2-5$
$\Rightarrow y=1-2-5$
$\Rightarrow y=-6$
Therefore, the vertex$(x,y)=(\dfrac{1}{2},-6)$.
Thus,
$\Rightarrow Vertex=(\dfrac{1}{2},-6)$, ‘a’ = 4
Now, substituting these values in the vertex equation form, we obtain
$y=a{{\left( x-h \right)}^{2}}+k$
$y=4{{\left( x-\dfrac{1}{2} \right)}^{2}}+\left( -6 \right)$
Simplifying the above, we get
$y={{\left( 4x-2 \right)}^{2}}-6$
Thus, vertex form of an equation is,
$\Rightarrow y={{\left( 4x-2 \right)}^{2}}-6$
$\Rightarrow Vertex=(\dfrac{1}{2},-6)$,
$\Rightarrow line\ of\ symmetry=x-coordinate=\dfrac{1}{2}$

Hence, the equation $y=4{{x}^{2}}-4x-5$ in vertex form is $y={{\left( 4x-2 \right)}^{2}}-6$ having $Vertex=(\dfrac{1}{2},-6)$ and $line\ of\ symmetry=x-coordinate=\dfrac{1}{2}$.

Note: Students should be very well aware of the vertex form of the equation. Also, they should know about the concept of converting the given equation into vertex form. They should also be very careful while doing the calculations to avoid making errors. Students should also know to find the coordinates of the vertex.