Chapter 9, Problem 6MCQ

To determine

Find the coordinates of the vertex. Find the vertex is maximum or minimum.

Answer to Problem 6MCQ

Vertex is minimum and the coordinates of the vertex is $(\frac{7}{2},-\frac{25}{4})$

Explanation of Solution

Given:

The equation: $y=x^2-7x+6$

Concept Used:

The quadratic equation: $y=a{x}^2+bx+c$

X − Coordinate of the vertex: $x=-\frac{b}{2a}$

Rule: the coefficient of $x^2$ is positive, the graph is upward open and vertex is minimum.

Calculation:

The quadratic equation: $y=a{x}^2+bx+c$

The x coordinate of the vertex: $x=-\frac{b}{2a}$

  $\begin{array}{l}y=x^2-7x+6\\ a=1,b=-7\\ x=-\frac{b}{2a}=-\frac{-7}{2·1}=\frac{7}{2}\\ x=\frac{7}{2}\end{array}$ x coordinate of the vertex is $x=\frac{7}{2}$

Now find the y − coordinate by substituting the value of $x=\frac{7}{2}$ in the equation $y=x^2-7x+6$

  $\begin{array}{l}y=x^2-7x+6\\ $ y={(\frac{7}{2})}^2-7·\frac{7}{2}+6[x=\frac{7}{2}]$y=\frac{49}{4}-\frac{49}{2}+\frac{6}{1}=\frac{49}{4}-\frac{98}{4}+\frac{24}{4}=\frac{49-98+24}{4}=-\frac{25}{4}\\ y=-\frac{25}{4}\end{array}$

Vertex of the equation $y=x^2-7x+6$ is $(\frac{7}{2},-\frac{25}{4})$

Rule: the coefficient of $x^2$ is positive, the graph is upward open and vertex is minimum.

Thus, Vertex is minimum and the coordinates of the vertex is $(\frac{7}{2},-\frac{25}{4})$