Chapter 2.1, Problem 30E

Solution

Find the vertex and axis of the graph to the function $f\left(x\right)=6-2x+4x^2$ . Rewrite the equation for the function in vertex form.

The vertex of the parabola is $\left(\frac{1}{4},\frac{23}{4}\right)$ .

And the axis of the parabola is given by $x=\frac{1}{4}$ .

Vertex form of the parabola is $f\left(x\right)=4{\left(x-\frac{1}{4}\right)}^2+\frac{23}{4}$ .

Given:

  $f\left(x\right)=6-2x+4x^2$ .

Concept Used:

• The vertex form of a parabola is given by $y=a{(x-h)}^2+k$
• The axis of the graph of the parabola is $x=h$

Calculation:

The given function is $f\left(x\right)=6-2x+4x^2$ .

Rewrite this in standard form $f\left(x\right)=4x^2-2x+6$

Use completing square method to write to write the function in vertex form

  $\begin{array}{c}f\left(x\right)=4x^2-2x+6\\ =4\left(x^2-\frac{x}{2}+\frac{3}{2}\right)\\ =4\left(x^2-\frac{x}{2}+\frac{3}{2}+{\left(-\frac{1}{4}\right)}^2-{\left(-\frac{1}{4}\right)}^2\right)\left(\text{Add and subtract }{\left(-\frac{1}{4}\right)}^2\right)\\ =4\left({\left(x-\frac{1}{4}\right)}^2+\frac{3}{2}-{\left(-\frac{1}{4}\right)}^2\right)\\ =4{\left(x-\frac{1}{4}\right)}^2+\frac{23}{4}\end{array}$

So, the vertex form of the parabola is $f\left(x\right)=4{\left(x-\frac{1}{4}\right)}^2+\frac{23}{4}$

Comparing this function with the vertex form $y=a{(x-h)}^2+k$

  $h=\frac{1}{4},k=\frac{23}{4}$

Hence, the vertex of the parabola is given by

  $\begin{array}{l}\left(h,k\right)\\ =\left(\frac{1}{4},\frac{23}{4}\right)\end{array}$ .

And the axis of the parabola is given by

  $\begin{array}{c}x=h\\ x=\frac{1}{4}\end{array}$