Chapter 9.1, Problem 65S

To determine

To find: The vertex form of the quadratic equation, identify the vertex, axis of symmetry and direction of opening.

Answer to Problem 65S

The vertex form of the equation is $y={\left(x-3\right)}^2-8$ , the vertex is $\left(h,k\right)=\left(3,-8\right)$ and the axis of symmetry is $x=3$ . As $a=1$ is negative the positive open upwards.

Explanation of Solution

Given:

The given equation is $y=x^2-6x+1$ .

Calculation:

Consider the given equation is,

  $y=x^2-6x+1$

Write the above equation in the vertex form as,

  $\begin{array}{c}y=x^2-6x+1\\ =\left(x^2-6+9\right)-9+1\\ ={\left(x-3\right)}^2-8\end{array}$

Consider the general vertex form is,

  $y=a{\left(x-h\right)}^2+k$

Here, $\left(h,k\right)$ are the vertex, $x$ is the axis of symmetry.

Then, the vertex form is $\left(h,k\right)=\left(3,-8\right)$

The axis of symmetry is $x=3$ .

The value of $a=1$ . As it is positive the parabola open upwards.

Thus, vertex form of the equation is $y={\left(x-3\right)}^2-8$ , the vertex is $\left(h,k\right)=\left(3,-8\right)$ and the axis of symmetry is $x=3$ . As $a=1$ is negative the positive open upwards.