91Ó°ÊÓ

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=3x^2-7x+2$ to the nearest hundredth.

The maximum or minimum value of a quadratic function lies at the vertex of the graph.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$:

if $a>0$, it is an upwards opening parabola and so has a minimum value

if a<0, it is a downwards opening parabola and so has a maximum value.

So the given quadratic equation has a minimum value.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$, the x-coordinate of the vertex is given by $x=-\frac{b}{2a}$

Here for the given equation, $a=3,b=-7$

Plugging the values in the equation:

$\begin{array}{l}x=−\frac{b}{2a}\\ x=−\frac{\left(−7\right)}{2\left(3\right)}\\ x=\frac{7}{6}\\ x≈1.17\end{array}$

Hence the x-coordinate of the vertex is x=1.17.

So, the y-coordinate of the vertex can be obtained by plugging the value in the equation.

Plugging $x=\frac{7}{6}$ in the equation:

$\begin{array}{l}f\left(x\right)=3x^2−7x+2\\ f\left(\frac{7}{6}\right)=3{\left(\frac{7}{6}\right)}^2−7\left(\frac{7}{6}\right)+2\\ f\left(\frac{7}{6}\right)=3\left(\frac{49}{36}\right)−7\left(\frac{7}{6}\right)+2\\ f\left(\frac{7}{6}\right)=\frac{49}{12}−\frac{98}{12}+\frac{24}{12}\\ f\left(\frac{7}{6}\right)=\frac{49−98+24}{12}\\ f\left(\frac{7}{6}\right)=−\frac{25}{12}\\ f\left(\frac{7}{6}\right)≈−2.08\end{array}$

Hence the coordinates of the vertex is $\left(1.17,-2.08\right)$.

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=3x^2-7x+2$ to the nearest hundredth is $\left(1.17,-2.08\right)$.