Chapter 2.1, Problem 27E

Solution

Find the vertex and axis of the graph to the function $f\left(x\right)=3x^2+5x-4$ . Rewrite the equation for the function in vertex form.

The vertex of the parabola is $\left(-\frac{5}{6},-\frac{73}{12}\right)$ .

And the axis of the parabola is given by $x=-\frac{5}{6}$ .

Vertex form of the parabola is $f\left(x\right)=3{\left(x+\frac{5}{6}\right)}^2-\frac{73}{12}$ .

Given:

  $f\left(x\right)=3x^2+5x-4$ .

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 $f\left(x\right)=3x^2+5x-4$ .

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

  $\begin{array}{c}f\left(x\right)=3x^2+5x-4\\ =3\left(x^2+\frac{5x}{3}-\frac{4}{3}\right)\\ =3\left(x^2+\frac{5x}{3}-\frac{4}{3}+{\left(\frac{5}{6}\right)}^2-{\left(\frac{5}{6}\right)}^2\right)\left(\text{Add and subtract }{\left(\frac{5}{6}\right)}^2\right)\\ =3\left({\left(x+\frac{5}{6}\right)}^2-\frac{4}{3}-{\left(\frac{5}{6}\right)}^2\right)\\ =3{\left(x+\frac{5}{6}\right)}^2-\frac{73}{12}\end{array}$

So, the vertex form of the parabola is $f\left(x\right)=3{\left(x+\frac{5}{6}\right)}^2-\frac{73}{12}$

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

  $h=-\frac{5}{6},k=-\frac{73}{12}$

Hence, the vertex of the parabola is given by

  $\begin{array}{l}\left(h,k\right)\\ =\left(-\frac{5}{6},-\frac{73}{12}\right)\end{array}$ .

And the axis of the parabola is given by

  $\begin{array}{c}x=h\\ x=-\frac{5}{6}\end{array}$