Chapter 2.1, Problem 4E

To determine

To find:Whether the graph of a quadratic function has relative minimum value at its vertex.

Answer to Problem 4E

The quadratic function $f\left(x\right)=-3x^2+5x+2$ does not have its relative minimum at its vertex.

Explanation of Solution

Given data: $y$

The given quadratic function is $f\left(x\right)=-3x^2+5x+2$

Method/Formula used:

The quadratic function $f\left(x\right)=a{x}^2+bx+c$ represents a parabola whose vertex is $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ . If $a>0$ then f has relative minimum at $x=-\frac{b}{2a}$ , and if $a<0$ then f has maximum at $x=-\frac{b}{2a}$ ,

Calculation:

The given quadratic function is $f\left(x\right)=-3x^2+5x+2$ .

Comparing given function with $f\left(x\right)=a{x}^2+bx+c$ , we have $a=-3<0$ , $b=5$ , $c=2$ .

Now, $-\frac{b}{2a}=-\frac{5}{2\cdot (-3)}=\frac{5}{6}$ , and $f\left(\frac{5}{6}\right)=-3{\left(\frac{5}{6}\right)}^2+5\left(\frac{5}{6}\right)+2=-\frac{75}{36}+\frac{25}{6}+2$

  $\Rightarrow f\left(\frac{5}{6}\right)=\frac{-75+150+72}{36}=\frac{147}{36}=\frac{49}{12}$

The vertex of the parabola is $\left(\frac{5}{6},f\left(\frac{5}{6}\right)\right)\equiv \left(\frac{5}{6},\frac{49}{12}\right)$ , shown in Figure-1 here.

Since, $a=-3<0$ , the parabola has relative maxima at the vertex $\left(\frac{5}{6},\frac{49}{12}\right)=\left(0.833,4.083\right)$ .

Conclusion:

The quadratic function

  $f\left(x\right)=-3x^2+5x+2$ does not have its relative minimum at its vertex.