Question

How do you find the vertex of $y=4{{x}^{2}}+8x+7?$

Answer 1

The vertex of an equation of the form $y=a{{x}^{2}}+bx+c$ can be found by finding the values of $h$ and $k.$ Here, $h$ is the $x-$coordinate of the vertex and $k$ is the $y-$coordinate of the vertex. And we can find these values by $h=\dfrac{-b}{2a}$ and $k$ by substituting $h$ for $x$ in the equation.

Complete step by step solution:
Consider the equation of the form $y=a{{x}^{2}}+bx+c.$
Suppose that $h$ is the $x-$coordinate of the vertex and $k$ is the corresponding $y-$coordinate of the vertex.
The vertex $\left( h,k \right)$ of the above equation can be found by the equation $h=\dfrac{-b}{2a}$ where $a$ is the coefficient of ${{x}^{2}},$ $b$ is the coefficient of $x.$ $k$ can be found by substituting the value of $h$ for $x$ in the equation.
Let us consider the given equation $y=4{{x}^{2}}+8x+7.$
We are asked to find the vertex of the given equation.
So, we will write the coefficient of each term first.
The leading coefficient, the coefficient of the term ${{x}^{2}},$ is $4.$ That is, $a=4.$
The coefficient $b$ of the term $x$ is $8.$ That is $b=8.$
When we substitute these values in $h=\dfrac{-b}{2a},$ we will get the $x-$coordinate as $h=\dfrac{-8}{2\cdot \left( 4 \right)}.$
We will get $h=\dfrac{-8}{8}=-1.$
Thus, the $x-$coordinate of the vertex is $h=-1.$
We will apply $h=-1$ in the given equation to get the $y-$coordinate $k$ of the vertex.
So, $k=4{{\left( -1 \right)}^{2}}+8\left( -1 \right)+7.$
Since ${{\left( -1 \right)}^{2}}=1,$ we will get $k=4\cdot 1+8\left( -1 \right)+7.$
That is, $k=4\cdot 1-8\cdot 1+7.$
Therefore the $y-$coordinate of the vertex of the given equation is $k=4-8+7=11-8=3.$
Hence the vertex of the given equation is $\left( h,k \right)=\left( -1,3 \right).$

Note: The $y-$coordinate $k$ can be obtained using another equation given by $k=c-\dfrac{{{b}^{2}}}{4a}.$ $c$ is the constant term in the equation, $c=7.$ Now we will get the $y-$coordinate $k=7-\dfrac{{{8}^{2}}}{4\cdot 4}=7-\dfrac{64}{16}=7-4=3.$