Question

How do you find the vertex of $ y = 2{x^2} - 6x + 1? $

Answer 1

Hint: As we know that the above given equation is a quadratic equation and a quadratic equation is also called a parabola. The vertex of a quadratic equation is nothing but the highest or lowest point of the quadratic equation. We know that the equation of the vertex form of the quadratic equation is $ y = a{(x - h)^2} + k $ . Here we have to change the quadratic equation into the vertex form, so we have to find the coordinates of the vertex and then substitute them in the formula with the help of $ \dfrac{{ - b}}{{2a}} $ .

Complete step-by-step answer:
As per the question we have the equation: $ y = 2{x^2} - 6x + 1 $ Here we will find the values of the variable $ a $ and $ b $ by comparing the equation. The form of the quadratic equation is $ y = a{x^2} + bx + c $ ,
So we have $ a = 2,b = - 6 $ and $ c = 1 $ . Now we will find the coordinates of the vertex which is $ (h,k) $ where $ h $ represents the x coordinate and $ k $ represents the y coordinate. And we know that the formula to the $ h $ coordinate is
$ \dfrac{{ - b}}{{2a}} $
By substituting the values we get
$ h = \dfrac{{ - ( - 6)}}{{2 \times 2}} $ , So we have
 $ h = \dfrac{3}{2} $ .
To find the value of $ k $ , we will substitute the values in the equation as $ x $ is represented by $ h $ . So we get the equation: $ y = 2{x^2} - 6x + 1 $ , by putting values we have:
 $ k = 2{\left( {\dfrac{3}{2}} \right)^2} - 6 \times \dfrac{3}{2} + 1 $ .
We will solve this now:
 $ k = 2 \times \dfrac{9}{4} - 9 + 1\\
 \Rightarrow k = \dfrac{{9 - 18 + 2}}{2} $ .
Therefore the value of $ k = \dfrac{{ - 7}}{2} $ .
Therefore we have $ h = \dfrac{3}{2} $ and $ k = \dfrac{{ - 7}}{2} $ . We will put these values in the vertex form, So we have coordinates of the function.
Hence the vertex is $ \left( {\dfrac{3}{2},\dfrac{{ - 7}}{2}} \right) $ .
So, the correct answer is “ $ \left( {\dfrac{3}{2},\dfrac{{ - 7}}{2}} \right) $ ”.

Note: We can solve the above equation by the method of completing the squares. We should be careful while finding the vertex of the quadratic equation while taking the constant terms and their respective positive and negative signs. We should also have the knowledge of quadratic equations, the vertex and their forms before solving this kind of equation.