
How do you find the vertex of $y = 4{x^2} - 6x - 3$ ?
Answer
545.4k+ views
Hint:
We have given an equation as $y = 4{x^2} - 6x - 3$ , which is a parabolic equation. The standard parabolic equation is always represented as $y = a{x^2} + bx + c$ and the expression $ - \dfrac{b}{{2a}}$ gives the x-coordinate, and then substitute x-coordinate in the original equation to obtain y-coordinate.
Complete step by step solution:
We have given an equation of parabola $y = 4{x^2} - 6x - 3$, (original equation)
We have to find the vertex of parabola ,
We know that the standard parabolic equation is always represented as $y = a{x^2} + bx + c$ and the expression $ - \dfrac{b}{{2a}}$ gives the x-coordinate , and when we substitute x-coordinate in the original equation we get the y-coordinate.
Now, compare the original equation with the standard parabolic equation, we will get,
The x-coordinate of the vertex , \[ - \dfrac{b}{{2a}} = - \dfrac{{( - 6)}}{{2(4)}}\]
$ = \dfrac{3}{4}$
Now substitute $x = \dfrac{3}{4}$ in the original equation to get y-coordinate as,
$ \Rightarrow y = 4{x^2} - 6x - 3$
$ \Rightarrow y = 4{\left( {\dfrac{3}{4}} \right)^2} - 6\left( {\dfrac{3}{4}} \right) - 3$
$ = \dfrac{{36}}{{16}} - \dfrac{{18}}{4} - 3$
$ = \dfrac{9}{4} - \dfrac{{18}}{4} - \dfrac{{12}}{4}$
$ = - \dfrac{{21}}{4}$
Therefore, the required vertex of the parabola is $\left( {\dfrac{3}{4}, - \dfrac{{21}}{4}} \right)$.
Note:
The equation for a parabola can also be written in vertex form as $y = a{(x - h)^2} + k$ where $(h,k)$ is the vertex of parabola. The point where a parabola has zero gradient is known as the vertex of the parabola. We need to get ‘y’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘y’, we have to undo the mathematical operations such as addition, subtraction, multiplication, and division that have been done to the variables. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
We have given an equation as $y = 4{x^2} - 6x - 3$ , which is a parabolic equation. The standard parabolic equation is always represented as $y = a{x^2} + bx + c$ and the expression $ - \dfrac{b}{{2a}}$ gives the x-coordinate, and then substitute x-coordinate in the original equation to obtain y-coordinate.
Complete step by step solution:
We have given an equation of parabola $y = 4{x^2} - 6x - 3$, (original equation)
We have to find the vertex of parabola ,
We know that the standard parabolic equation is always represented as $y = a{x^2} + bx + c$ and the expression $ - \dfrac{b}{{2a}}$ gives the x-coordinate , and when we substitute x-coordinate in the original equation we get the y-coordinate.
Now, compare the original equation with the standard parabolic equation, we will get,
The x-coordinate of the vertex , \[ - \dfrac{b}{{2a}} = - \dfrac{{( - 6)}}{{2(4)}}\]
$ = \dfrac{3}{4}$
Now substitute $x = \dfrac{3}{4}$ in the original equation to get y-coordinate as,
$ \Rightarrow y = 4{x^2} - 6x - 3$
$ \Rightarrow y = 4{\left( {\dfrac{3}{4}} \right)^2} - 6\left( {\dfrac{3}{4}} \right) - 3$
$ = \dfrac{{36}}{{16}} - \dfrac{{18}}{4} - 3$
$ = \dfrac{9}{4} - \dfrac{{18}}{4} - \dfrac{{12}}{4}$
$ = - \dfrac{{21}}{4}$
Therefore, the required vertex of the parabola is $\left( {\dfrac{3}{4}, - \dfrac{{21}}{4}} \right)$.
Note:
The equation for a parabola can also be written in vertex form as $y = a{(x - h)^2} + k$ where $(h,k)$ is the vertex of parabola. The point where a parabola has zero gradient is known as the vertex of the parabola. We need to get ‘y’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘y’, we have to undo the mathematical operations such as addition, subtraction, multiplication, and division that have been done to the variables. Use addition or subtraction properties of equality to gather variable terms on one side of the equation and constant on the other side of the equation. Use the multiplication or division properties of equality to form the coefficient of the variable term equivalent to one.
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