Chapter 9, Problem 6MCQ

To determine

To find the vertex of the function.

Answer to Problem 6MCQ

The function has a minimum value at its vertex $(2.5,-2.25)$ .

Explanation of Solution

Given information :

The function provided is :- $y=x^2-5x+4$

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

1. Highest point (if $a<0$) and be called the maximum. Here, the graph opens downwards.
2. Lowest point (if $a>0$) and be called the minimum. Here, the graph opens upwards.

Here, a is greater than zero, hence the parabola opens upwards, and has a minimum value.

A parabola always points to infinity, either negative or positive.

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

Substituting the values from the given equation.

  $x=-\frac{(-5)}{2\times 1}$

Simplifying.

  $x=\frac{5}{2}$

This corresponds to the x -coordinate of the vertex

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

For example, to compute the value of ‘y’ when $x=\frac{5}{2}$ , substitute the value of ‘x’ in equation $y=x^2-5x+4$

On substituting the value of the ‘x’ as $x=\frac{5}{2}$ , it is,

  $y={(\frac{5}{2})}^2-5(\frac{5}{2})+4$

On simplifying this,

  $=\frac{25}{4}-\frac{25}{2}+4$

Finally, the value of ‘y’ is

  $=-\frac{9}{4}$

The vertex of the function is $(2.5,-2.25)$ .

Therefore, the function $y=x^2-5x+4$ has a minimum value of $-2.25$ .