Chapter 8.1, Problem 8QR

Solution

To find: The vertex and axis of the graph of f and describe how the graph f can be obtained from the graph of $g(x)=x^2$ and then graph f .

Vertex: $\left(3,19\right)$

Axis of graph: $x=3$

To obtain $f\left(x\right)$ : $g(x)=x^2$: Shift 3 units to the right, stretch 2 times along the y -axis, reflect about x -axis and then move 19 units up.

  

Given: $f(x)=-2x^2+12x+1$

Concept used:

The vertex form of parabola is given by:-

  $y=a{(x-h)}^2+k$ , where $\left(h,k\right)$ is the vertex.

Th axis of the graph is $x=h.$

If $\left|a\right|>1$ , then the graph is said to be vertically stretched by factor $‘a’$ .

If $0<\left|a\right|<1$ , then the graph is said to be vertically compressed by factor $‘a’$ .

  $-a$ suggest reflection across x -axis.

Calculation:

Use completing the square method to write the given equation in vertex form.

Take out -2 as common from the first two terms.

  $f(x)=-2\left(x^2-6x\right)+1$

Compare the expression inside the bracket with $x^2-bx$ to get the value of b .

Here $b=-6$

Add and subtract the value of ${\left(\frac{b}{2}\right)}^2$ .

  $\begin{array}{c}f(x)=-2\left(x^2-6x+{\left(\frac{6}{2}\right)}^2-{\left(\frac{6}{2}\right)}^2\right)+1\\ =-2\left(x^2-6x+9-9\right)+1\\ =-2\left({\left(x-3\right)}^2-9\right)+1\left[a^2-2ab+b^2={\left(a-b\right)}^2\right]\\ =-2{\left(x-3\right)}^2+18+1\\ =-2{\left(x-3\right)}^2+19\end{array}$

Comparing the given equation with the standard equation, to get the vertex of the parabola which is $\left(3,19\right)$ and the axis of the graph will be $x=3$ .

To obtain the graph of $f\left(x\right)$ from the graph of $g(x)=x^2$ : Shift 3 units to the right, stretch 2 times along the y -axis, reflect about x -axis and then move 19 units up.

The graph of $f\left(x\right)$ is shown: