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Question: How to find the vertex of a parabola without its graphing?...

How to find the vertex of a parabola without its graphing?

Explanation

Solution

The above given problem is a very simple problem of coordinate geometry. The given question is related to the concept of parabola and for solving these types of questions we first need to understand the different forms of parabolas. There are four types of parabola and for each of them, the corresponding vertex and foci are as follows:

EquationVertexFoci
y2=4ax{y^2} = 4ax(0,0)\left( {0,0} \right)(a,0)\left( {a,0} \right)
y2=4ax{y^2} = - 4ax(0,0)\left( {0,0} \right)(a,0)\left( { - a,0} \right)
x2=4by{x^2} = 4by(0,0)\left( {0,0} \right)(0,b)\left( {0,b} \right)
x2=4by{x^2} = - 4by(0,0)\left( {0,0} \right)(0,b)\left( {0, - b} \right)

Complete step by step solution:
We are supposed to figure out how to find the vertex of a parabola without its graphing.
Let us assume that the parabola has not been rotated. The vertex of the parabola occurs at the point where the derivative of its function is equal to zero.
First of all, we have to determine the general form of the derivative, which should be a linear function in the terms of x.
Next, we have to set the general form of the derivative to zero and then solve for the value of x.
As the next step, we have to substitute the value obtained from x in the parabola expression back and get the y component of the vertex coordinate.
That’s how we find the vertex of a parabola without graphing.

Note:
The given question was an easy one. Students should be familiar with the concept of parabola and have full knowledge of different types of parabolas. An example of the above explained process is: if a parabola is given by y=f(x)=3x2+12x+7y = f\left( x \right) = 3{x^2} + 12x + 7. Then,
f(x)=6x+12f'\left( x \right) = 6x + 12, the vertex occurs where derivative is equal to zero,
6x+12=0 x=2 y=3(2)2+12(2)+7 y=5  \Rightarrow 6x + 12 = 0 \\\ \Rightarrow x = - 2 \\\ y = 3{\left( { - 2} \right)^2} + 12\left( { - 2} \right) + 7 \\\ y = - 5 \\\
Therefore, the vertex is at (2,5)\left( { - 2, - 5} \right).