Chapter 9.1, Problem 66S

To determine

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

Answer to Problem 66S

The vertex form of the equation is $y=-2{\left(x-5\right)}^2+15$ , the vertex is $\left(h,k\right)=\left(5,15\right)$ and the axis of symmetry is $x=5$ . As $a=1$ is negative the positive open upwards.

Explanation of Solution

Given:

The given equation is $y=-2x^2+20x-35$ .

Calculation:

Consider the given equation is,

  $y=-2x^2+20x-35$

Write the above equation in the vertex form as,

  $\begin{array}{c}y=-2x^2+20x-35\\ =-2\left(x^2-10x+25\right)+50-35\\ =-2{\left(x-5\right)}^2+15\end{array}$

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 $\left(h,k\right)=\left(5,15\right)$

The axis of symmetry is $x=5$ .

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

Thus, vertex form of the equation is $y=-2{\left(x-5\right)}^2+15$ , the vertex is $\left(h,k\right)=\left(5,15\right)$ and the axis of symmetry is $x=5$ . As $a=1$ is negative the positive open upwards.