Q60. - Algebra 2

Question

Find the coordinates of the maximum or minimum value of each quadratic equation to the nearest hundredth.

 

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

Step-by-Step Solution

Answer

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=2x^2-3x+2$ to the nearest hundredth is $\left(0.75,0.875\right)$.

Step 1. Given Information.

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

Step 2. Explanation .

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=2,b=-3

Plugging the values in the equation:

 $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(-3\right)}{2\left(2\right)}\\ x=\frac{3}{4}\\ x=0.75\end{array}$

Hence the x-coordinate of the vertex is  $x=0.75$.

 

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

Plugging $x=0.75$  in the equation:

 $\begin{array}{l}f\left(x\right)=2x^2-3x+2\\ f\left(0.75\right)=2{\left(0.75\right)}^2-3\left(0.75\right)+2\\ f\left(0.75\right)=2\left(0.5625\right)-3\left(0.75\right)+2\\ f\left(0.75\right)=1.125-2.25+2\\ f\left(0.75\right)=0.875\end{array}$

Hence the coordinates of the vertex is $\left(0.75,0.875\right)$.

Step 3. Conclusion .

The coordinates of the minimum value of the quadratic equation $f\left(x\right)=2x^2-3x+2$ to the nearest hundredth is $\left(0.75,0.875\right)$.